Three Dimensional Geometry jee mains questions with solutions

Q 1 :

Let the point, on the line passing through the points P(1, –2, 3) and Q(5, –4, 7), farther from the origin and at a distance of 9 units from the point P, be $(α, β, γ)$ . Then $α^{2} + β^{2} + γ^{2}$ is equal to [2024]

160
165
150
155

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(4)

Line through PQ is given by

$\frac{x - 1}{4} = \frac{y + 2}{- 2} = \frac{z - 3}{4} = λ$

Any point R on PQ will be $R (4 λ + 1, - 2 λ - 2, 4 λ + 3)$

Since, PR = 9 units $\Rightarrow {(P R)}^{2} = 81$

$\Rightarrow {(4 λ + 1 - 1)}^{2} + {(- 2 λ - 2 + 2)}^{2} + {(4 λ + 3 - 3)}^{2} = 81$

$\Rightarrow 16 λ^{2} + 4 λ^{2} + 16 λ^{2} = 81$

$\Rightarrow 36 λ^{2} = 81 \Rightarrow λ^{2} = \frac{9}{4} \Rightarrow λ = \pm \frac{3}{2}$

$∴$ R can be (7, –5, 9) or (–5, 1, –3)

Distance from origin of the points be $\sqrt{49 + 25 + 81}$ and $\sqrt{25 + 1 + 9}$ i.e., $\sqrt{155}, \sqrt{35}$

$∴$ Distance of (7, –5, 9) is farthest from origin

$\Rightarrow$ $(α, β, γ) \equiv (7, - 5, 9)$

Hence $α^{2} + β^{2} + γ^{2} = 7^{2} + {(- 5)}^{2} + 9^{2} = 155$ .

Q 2 :

Let P be the point of intersection of the lines $\frac{x - 2}{1} = \frac{y - 4}{5} = \frac{z - 2}{1}$ and $\frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 3}{2}$ . Then, the shortest distance of P from the line 4x = 2y = z is [2024]

$\frac{\sqrt{14}}{7}$
$\frac{5 \sqrt{14}}{7}$
$\frac{6 \sqrt{14}}{7}$
$\frac{3 \sqrt{14}}{7}$

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2

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4

(4)

$L_{1} : \frac{x - 2}{1} = \frac{y - 4}{5} = \frac{z - 2}{1} = λ (say)$

$L_{2} : \frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 3}{2} = μ (say)$

For point of intersection P,

$(λ + 2, 5 λ + 4, λ + 2) = (2 μ + 3, 3 μ + 2, 2 μ + 3)$

$\Rightarrow λ + 2 = 2 μ + 3, 5 λ + 4 = 3 μ + 2, λ + 2 = 2 μ + 3$

$\Rightarrow λ = - 1, μ = - 1$

$∴$ Point P is (1, –1, 1)

Distance of point P from $L_{3}$ : 4x = 2y = z

$L_{3} : \frac{x}{\frac{1}{4}} = \frac{y}{\frac{1}{2}} = \frac{z}{1}$

Any point on $L_{3}$ be $R (\frac{α}{4}, \frac{α}{2}, α)$

D.r.'s of $P R : (\frac{α}{4} - 1, \frac{α}{2} + 1, α - 1) ∵ P R ⊥ (\frac{1}{4}, \frac{1}{2}, 1)$

$\Rightarrow (\frac{α}{4} - 1) \frac{1}{4} + \frac{1}{2} (\frac{α}{2} + 1) + (α - 1) = 0$

$\Rightarrow \frac{α}{16} - \frac{1}{4} + \frac{α}{4} + \frac{1}{2} + α - 1 = 0 \Rightarrow \frac{21}{16} α = \frac{3}{4} \Rightarrow α = \frac{4}{7}$

$∴ R (\frac{1}{7}, \frac{2}{7}, \frac{4}{7})$

Now, $R P = \sqrt{{(\frac{1}{7} - 1)}^{2} + {(\frac{2}{7} + 1)}^{2} + {(\frac{4}{7} - 1)}^{2}}$

$= \sqrt{\frac{36}{49} + \frac{81}{49} + \frac{9}{49}} = \frac{\sqrt{126}}{7} = \frac{3 \sqrt{14}}{7}$ .

Q 3 :

Let d be the distance of the point of intersection of the lines $\frac{x + 6}{3} = \frac{y}{2} = \frac{z + 1}{1}$ and $\frac{x - 7}{4} = \frac{y - 9}{3} = \frac{z - 4}{2}$ from the point(7, 8, 9). Then $d^{2} + 6$ is equal to [2024]

72
75
78
69

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(2)

Let $\frac{x + 6}{3} = \frac{y}{2} = \frac{z + 1}{1} = λ$

$\Rightarrow x = 3 λ - 6, y = 2 λ, z = λ - 1$

and let $\frac{x - 7}{4} = \frac{y - 9}{3} = \frac{z - 4}{2} = μ$

$\Rightarrow x = 4 μ + 7, y = 3 μ + 9, z = 2 μ + 4$

Since the given lines intersect so we have

$3 λ - 6 = 4 μ + 7, 2 λ = 3 μ + 9, λ - 1 = 2 μ + 4$

$\Rightarrow 3 λ - 4 μ = 13 a n d λ = 2 μ + 5$

$\Rightarrow 3 (2 μ + 5) - 4 μ = 13 \Rightarrow 2 μ = - 2$

$\Rightarrow μ = - 1 a n d λ = 3$

So, point of intersection is (3, 6, 2)

$∴ d = \sqrt{{(3 - 7)}^{2} + {(6 - 8)}^{2} + {(2 - 9)}^{2}}$

$= \sqrt{16 + 4 + 49} = \sqrt{69}$

$∴ d^{2} + 6 = 69 + 6 = 75$ .

Q 4 :

If the line $\frac{2 - x}{3} = \frac{3 y - 2}{4 λ + 1} = 4 - z$ makes a right angle with the line $\frac{x + 3}{3 μ} = \frac{1 - 2 y}{6} = \frac{5 - z}{7}$ , then $4 λ + 9 μ$ is equal to : [2024]

13
6
5
4

1

2

3

4

(2)

We have lines $\frac{2 - x}{3} = \frac{3 y - 2}{4 λ + 1} = 4 - z$ and $\frac{x + 3}{3 μ} = \frac{1 - 2 y}{6} = \frac{5 - z}{7}$ are perpendicular.

i.e., $\frac{x - 2}{- 3} = \frac{y - 2 / 3}{\frac{4 λ + 1}{3}} = \frac{z - 4}{- 1}$ and $\frac{x - (- 3)}{3 μ} = \frac{y - 1 / 2}{\frac{- 6}{2}} = \frac{z - 5}{- 7}$ are perpendicular.

$\Rightarrow - 3 (3 μ) + \frac{4 λ + 1}{3} (- 3) + (- 1) (- 7) = 0$

$\Rightarrow - 9 μ - 4 λ - 1 + 7 = 0$

$\Rightarrow - 9 μ - 4 λ + 6 = 0 \Rightarrow 4 λ + 9 μ = 6$ .

Q 5 :

Let $(α, β, γ)$ be the image of the point (8, 5, 7) in the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 2}{5}$ . Then $α + β + γ$ is equal to : [2024]

16
14
18
20

1

2

3

4

(2)

Let $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 2}{5} = λ$

$\Rightarrow x = 2 λ + 1, y = 3 λ - 1 and z = 5 λ + 2$

Clearly, $\vec{P Q} \cdot (2 \hat{i} + 3 \hat{j} + 5 \hat{k}) = 0$

$\Rightarrow (2 λ - 7) \cdot 2 + (3 λ - 6) \cdot 3 + (5 λ - 5) \cdot 5 = 0$

$\Rightarrow 38 λ = 57$

$\Rightarrow λ = \frac{57}{38} = \frac{3}{2}$

$∴$ $(4, \frac{7}{2}, \frac{19}{2})$ are the coordinates of Q.

Now, Q is the midpoint of PP'.

$∴ \frac{8 + α}{2} = 4, \frac{5 + β}{2} = \frac{7}{2}, \frac{7 + γ}{2} = \frac{19}{2}$

$\Rightarrow (α, β, γ) = (0, 2, 12)$

$∴ α + β + γ = 14$ .

Q 6 :

The shortest distance between the lines $\frac{x - 3}{2} = \frac{y + 15}{- 7} = \frac{z - 9}{5}$ and $\frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{- 3}$ is [2024]

$8 \sqrt{3}$
$5 \sqrt{3}$
$4 \sqrt{3}$
$6 \sqrt{3}$

1

2

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4

(3)

$L_{1} : \frac{x - 3}{2} = \frac{y + 15}{- 7} = \frac{z - 9}{5}$

$L_{2} : \frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{- 3}$

$\vec{a_{1}} = 3 \hat{i} - 15 \hat{j} + 9 \hat{k}, \vec{a_{2}} = - \hat{i} + \hat{j} + 9 \hat{k}$

$\vec{b_{1}} = 2 \hat{i} - 7 \hat{j} + 5 \hat{k}, \vec{b_{2}} = 2 \hat{i} + \hat{j} - 3 \hat{k}$

Shortest distance = $= \frac{| ({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2}) |}{| {\vec{b}}_{1} \times {\vec{b}}_{2} |}$

$= \frac{| (- 4 \hat{i} + 16 \hat{j}) \cdot | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & - 7 & 5 \\ 2 & 1 & - 3 \end{matrix} | |}{| {\vec{b}}_{1} \times {\vec{b}}_{2} |}$

$= \frac{| (- 4 \hat{i} + 16 \hat{j}) \cdot (16 \hat{i} + 16 \hat{j} + 16 \hat{k}) |}{| 16 \hat{i} + 16 \hat{j} + 16 \hat{k} |} = \frac{192}{16 \sqrt{3}} = 4 \sqrt{3}$ .

Q 7 :

Let $P (α, β, γ)$ be the image of the point Q(3, –3, 1) in the line $\frac{x - 0}{1} = \frac{y - 3}{1} = \frac{z - 1}{- 1}$ and R be the point (2, 5, –1). If the area of the triangle PQR is $λ$ and $λ^{2} = 14 K$ , then K is equal to : [2024]

81
72
36
18

1

2

3

4

(1)

$L : \frac{x - 0}{1} = \frac{y - 3}{1} = \frac{z - 1}{- 1} = μ$ be the given line.

Any point on L is given by $M (μ, μ + 3, - μ + 1)$

Direction ratios of $Q M = (μ - 3, μ + 6, - μ)$

Direction ratios of L = (1, 1, –1)

Now, QM $⊥$ L

So, we have, $1 \cdot (μ - 3) + 1 \cdot (μ + 6) - 1 \cdot (- μ) = 0$

$\Rightarrow 3 μ + 3 = 0$

$\Rightarrow μ = - 1$

So, M(–1, 2, 2) is mid point of PQ.

$\Rightarrow \frac{α + 3}{2} = - 1, \frac{β - 3}{2} = 2, \frac{γ + 1}{2} = 2$

$\Rightarrow P (α, β, γ) \equiv (- 5, 7, 3)$

Area of $△ P Q R = \frac{1}{2} | \vec{P Q} \times \vec{Q R} |$

$\Rightarrow λ = \frac{1}{2}$ $| | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 8 & - 10 & - 2 \\ - 1 & 8 & - 2 \end{matrix} | |$

$\Rightarrow λ = \frac{1}{2}$ $| 36 \hat{i} + 18 \hat{j} + 54 \hat{k} |$

$\Rightarrow λ^{2} = \frac{1}{4} (4536)$

$\Rightarrow λ^{2} = 1134 = 14 K \Rightarrow k = \frac{1134}{14} = 81$ .

Q 8 :

If the shortest distance between the lines

$L_{1} : \vec{r} = (2 + λ) \hat{i} + (1 - 3 λ) \hat{j} + (3 + 4 λ) \hat{k}, λ \in R$

$L_{2} : \vec{r} = 2 (1 + μ) \hat{i} + 3 (1 + μ) \hat{j} + (5 + μ) \hat{k}, μ \in R$

is $\frac{m}{\sqrt{n}}$ , where gcd(m, n) = 1, then the value of m + n equals [2024]

384
387
390
377

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2

3

4

(2)

$L_{1} : \frac{x - 2}{1} = \frac{y - 1}{- 3} = \frac{z - 3}{4}$

$L_{2} : \frac{x - 2}{2} = \frac{y - 3}{3} = \frac{z - 5}{1}$

$∴ a_{1} = 2 \hat{i} + \hat{j} + 3 \hat{k}, a_{2} = 2 \hat{i} + 3 \hat{j} + 5 \hat{k}$

$n_{1} = \hat{i} - 3 \hat{j} + 4 \hat{k}, n_{2} = 2 \hat{i} + 3 \hat{j} + \hat{k}$

Shortest distance = $\frac{| ({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{n}}_{1} \times {\vec{n}}_{2}) |}{| {\vec{n}}_{1} \times {\vec{n}}_{2} |}$

$= \frac{| (2 \hat{j} + 2 \hat{k}) \cdot | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 3 & 4 \\ 2 & 3 & 1 \end{matrix} | |}{| {\vec{n}}_{1} \times {\vec{n}}_{2} |}$

$=$ $| \frac{(2 \hat{j} + 2 \hat{k}) \cdot (- 15 \hat{i} + 7 \hat{j} + 9 \hat{k})}{\sqrt{{(15)}^{2} + 7^{2} + 9^{2}}} |$ $= \frac{32}{\sqrt{355}} = \frac{m}{\sqrt{n}}$

On comparing, we get m = 32 and n = 355.

So, m + n = 32 + 355 = 387.

Q 9 :

If the shortest distance between the lines $\frac{x - λ}{2} = \frac{y - 4}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{4} = \frac{y - 4}{6} = \frac{z - 7}{8}$ is $\frac{13}{\sqrt{29}}$ , then a value of $λ$ is [2024]

$- \frac{13}{25}$
$\frac{13}{25}$
1
–1

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2

3

4

(3)

We have $\vec{a} = λ \hat{i} + 4 \hat{j} + 3 \hat{k}, \vec{b} = 2 \hat{i} + 4 \hat{j} + 7 \hat{k}$ and $\vec{p} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$

Shortest distance, $d =$ $| \frac{(\vec{b} - \vec{a}) \times \vec{p}}{| \vec{p} |} |$

$\vec{b} - \vec{a} = (2 - λ) \hat{i} + 4 \hat{k}$

$(\vec{b} - \vec{a}) \times \vec{p} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 - λ & 0 & 4 \\ 2 & 3 & 4 \end{matrix} |$

$= \hat{i} (- 12) - \hat{j} (8 - 4 λ - 8) + \hat{k} (6 - 3 λ)$

$= - 12 \hat{i} + 4 λ \hat{j} + 3 (2 - λ) \hat{k}$

$∴ d =$ $| \frac{\sqrt{144 + 16 λ^{2} + 9 {(2 - λ)}^{2}}}{\sqrt{4 + 9 + 16}} |$ $= \frac{13}{\sqrt{29}}$ (Given)

$\Rightarrow \sqrt{144 + 16 λ^{2} + 36 + 9 λ^{2} - 36 λ} = 13$

$\Rightarrow 25 λ^{2} - 36 λ + 180 = 169 \Rightarrow 25 λ^{2} - 36 λ + 11 = 0$

$\Rightarrow (25 λ - 11) (λ - 1) = 0 \Rightarrow λ = \frac{11}{25} or 1$ .

Q 10 :

The shortest distance between the lines $\frac{x - 3}{4} = \frac{y + 7}{- 11} = \frac{z - 1}{5}$ and $\frac{x - 5}{3} = \frac{y - 9}{- 6} = \frac{z + 2}{1}$ is [2024]

$\frac{179}{\sqrt{563}}$
$\frac{185}{\sqrt{563}}$
$\frac{187}{\sqrt{563}}$
$\frac{178}{\sqrt{563}}$

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2

3

4

(3)

Given lines can be written as

$\frac{x - 3}{4} = \frac{y - (- 7)}{- 11} = \frac{z - 1}{5}$ and $\frac{x - 5}{3} = \frac{y - 9}{- 6} = \frac{z - (- 2)}{1}$

${\vec{a}}_{1} = 3 \hat{i} - 7 \hat{j} + \hat{k}, {\vec{a}}_{2} = 5 \hat{i} + 9 \hat{j} - 2 \hat{k}, {\vec{b}}_{1} = 4 \hat{i} - 11 \hat{j} + 5 \hat{k}$

and ${\vec{b}}_{2} = 3 \hat{i} - 6 \hat{j} + \hat{k}$

$∴ {\vec{a}}_{2} - {\vec{a}}_{1} = 2 \hat{i} + 16 \hat{j} - 3 \hat{k}$ and ${\vec{b}}_{1} \times {\vec{b}}_{2} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & - 11 & 5 \\ 3 & - 6 & 1 \end{matrix} |$ $= 19 \hat{i} + 11 \hat{j} + 9 \hat{k}$

Shortest distance between two lines,

$d = \frac{| ({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2}) |}{| ({\vec{b}}_{1} \times {\vec{b}}_{2}) |} \Rightarrow d = \frac{| 38 + 176 - 27 |}{\sqrt{361 + 121 + 81}} = \frac{187}{\sqrt{563}}$ .

Q 11 :

Let the line L intersect the lines x – 2 = – y = z – 1, 2(x + 1) = 2(y –1) = z + 1 and be parallel to the line $\frac{x - 2}{3} = \frac{y - 1}{1} = \frac{z - 2}{2}$ . Then which of the following points lis on L? [2024]

$(- \frac{1}{3}, - 1, - 1)$
$(- \frac{1}{3}, 1, - 1)$
$(- \frac{1}{3}, - 1, 1)$
$(- \frac{1}{3}, 1, 1)$

1

2

3

4

(2)

$L_{1} : \frac{x - 2}{1} = \frac{y}{- 1} = \frac{z - 1}{1} = λ$

$L_{2} : \frac{x + 1}{(1 / 2)} = \frac{y - 1}{(1 / 2)} = \frac{z + 1}{1} = μ$

Any point on $L_{1}$ and $L_{2}$ will be of the form $(λ + 2, - λ, λ + 1)$ and $(\frac{μ}{2} - 1, \frac{μ}{2} + 1, μ - 1)$ respectively.

Now direction ratios of line L are given by

$< λ - \frac{μ}{2} + 3, - λ - \frac{μ}{2} - 1, λ - μ + 2 >$

Also L is parallel to line $\frac{x - 2}{3} = \frac{y - 1}{1} = \frac{z - 2}{2}$

$\Rightarrow \frac{λ - \frac{μ}{2} + 3}{3} = \frac{- λ - \frac{μ}{2} - 1}{1} = \frac{λ - μ + 2}{2}$

$\begin{matrix} \Rightarrow λ - \frac{μ}{2} + 3 = - 3 λ - \frac{3 μ}{2} - 3 . . . (i) \\ 2 (λ - \frac{μ}{2} + 3) = 3 (λ - μ + 2) . . . (ii) \end{matrix}} λ = \frac{- 4}{3}, μ = \frac{- 2}{3}$

$∴$ Points will be $(\frac{2}{3}, \frac{4}{3}, \frac{- 1}{3})$ and $(\frac{- 4}{3}, \frac{2}{3}, \frac{- 5}{3})$

$∴$ L is given by $\frac{x - \frac{2}{3}}{3} = \frac{y - \frac{4}{3}}{1} = \frac{z + \frac{1}{3}}{2}$

Point $(\frac{- 1}{3}, 1, - 1)$ will satisfy this line L.

Q 12 :

Consider the line L passing through the points (1, 2, 3) and (2, 3, 5). The distance of the point $(\frac{11}{3}, \frac{11}{3}, \frac{19}{3})$ from the line L along the line $\frac{3 x - 11}{2} = \frac{3 y - 11}{1} = \frac{3 z - 19}{2}$ is equal to [2024]

4
3
5
6

1

2

3

4

(2)

Equation of line L is given by

$\frac{x - 1}{2 - 1} = \frac{y - 2}{3 - 2} = \frac{z - 3}{5 - 3} = λ$

i.e., $\frac{x - 1}{1} = \frac{y - 2}{1} = \frac{z - 3}{2} = λ$

$∴$ Any point on L is given by $(λ + 1, λ + 2, 2 λ + 3)$

Also, any point of $L_{1} : \frac{3 x - 11}{2} = \frac{3 y - 11}{1} = \frac{3 z - 19}{2} = μ$

is given by $(\frac{2 μ + 11}{3}, \frac{μ + 11}{3}, \frac{2 μ + 19}{3})$

Now, $λ + 1 = \frac{2 μ + 11}{3}, λ + 2 = \frac{μ + 11}{3}, 2 λ + 3 = \frac{2 μ + 19}{3}$

$\Rightarrow 3 λ - 2 μ = 8 and 3 λ - μ = 5$

On solving these two equations, we get $λ = \frac{2}{3}$ and $μ = - 3$

Point on L is $(\frac{5}{3}, \frac{8}{3}, \frac{13}{3})$

Required distance = $\sqrt{{(\frac{11}{3} - \frac{5}{3})}^{2} + {(\frac{11}{3} - \frac{8}{3})}^{2} + {(\frac{19}{3} - \frac{13}{3})}^{2}}$

$= \sqrt{4 + 1 + 1} = 3$ units.

Q 13 :

If the shortest distance between the lines $\frac{x - λ}{- 2} = \frac{y - 2}{1} = \frac{z - 1}{1}$ and $\frac{x - \sqrt{3}}{1} = \frac{y - 1}{- 2} = \frac{z - 2}{1}$ is 1, then the sum of all possible values of $λ$ is: [2024]

$2 \sqrt{3}$
0
$3 \sqrt{3}$
$- 2 \sqrt{3}$

1

2

3

4

(1)

The shortest distance between the given lines is given by

$d =$ $| \frac{| \begin{matrix} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{matrix} |}{\sqrt{{(a_{1} b_{2} - a_{2} b_{1})}^{2} + {(b_{1} c_{2} - b_{2} c_{1})}^{2} + {(c_{1} a_{2} - c_{2} a_{1})}^{2}}} |$

$\Rightarrow 1 =$ $| \frac{| \begin{matrix} \sqrt{3} - λ & - 1 & 1 \\ - 2 & 1 & 1 \\ 1 & - 2 & 1 \end{matrix} |}{\sqrt{{(3)}^{2} + {(3)}^{2} + {(3)}^{2}}} |$

$\Rightarrow 1 = \frac{| (\sqrt{3} - λ) (3) + 1 (- 3) + 1 (3) |}{\sqrt{27}}$

$\Rightarrow 3 \sqrt{3} = 3 \sqrt{3} - 3 λ or - 3 \sqrt{3} = 3 \sqrt{3} - 3 λ$

$\Rightarrow λ = 0 or 6 \sqrt{3} = 3 λ \Rightarrow λ = 2 \sqrt{3}$

$∴$ The sum of possible values of $λ = 0 + 2 \sqrt{3} = 2 \sqrt{3}$ .

Q 14 :

Let P and Q be the points on the line $\frac{x + 3}{8} = \frac{y - 4}{2} = \frac{z + 1}{2}$ which are at a distance of 6 units from the point R(1, 2, 3). If the centroid of the triangle PQR is $(α, β, γ)$ , then $α^{2} + β^{2} + γ^{2}$ is : [2024]

18
24
26
36

1

2

3

4

(1)

The given equation of line is,

$\frac{x + 3}{8} = \frac{y - 4}{2} = \frac{z + 1}{2} = λ$ (say)

$∴$ Any point on line is $(8 λ - 3, 2 λ + 4, 2 λ - 1)$

Now, distance of this point from R(1, 2, 3)

$\Rightarrow {(8 λ - 4)}^{2} + {(2 λ + 2)}^{2} + {(2 λ - 4)}^{2} = 36$

$\Rightarrow 72 λ^{2} - 72 λ = 0 \Rightarrow λ (λ - 1) = 0$

$\Rightarrow λ = 0 or λ = 1$

So, coordinates of P = (–3, 4, –1) and coordinates of Q = (5, 6, 1).

Now, centroid of $△ P Q R$ = (1, 4, 1) $\Rightarrow α = 1, β = 4, γ = 1$

So, $α^{2} + β^{2} + γ^{2} = 18$ .

Q 15 :

If the mirror image of the point P(3, 4, 9) in the line $\frac{x - 1}{3} = \frac{y + 1}{2} = \frac{z - 2}{1}$ is $(α, β, γ)$ , then $14 (α + β + γ)$ is : [2024]

102
138
132
108

1

2

3

4

(4)

Given equation of line is

$\frac{x - 1}{3} = \frac{y + 1}{2} = \frac{z - 2}{1} = λ$ (say)

$∴$ Any point on line, say $m = (3 λ + 1, 2 λ - 1, λ + 2)$

Now, d.r.'s of $P M = < 3 λ - 2, 2 λ - 5, λ - 7 >$ and d.r.'s of given line = $< 3, 2, 1 > ∵ P M ⊥ line$

$∴ 3 (3 λ - 2) + 2 (2 λ - 5) + 1 (λ - 7) = 0$

$\Rightarrow 14 λ = 23 \Rightarrow λ = \frac{23}{14}$

Now, M is the mid point of P(3, 4, 9) and $Q (α, β, γ)$

$∴ \frac{83}{14} = \frac{3 + α}{2} \Rightarrow α = \frac{62}{7} \Rightarrow \frac{16}{7} = \frac{4 + β}{2} \Rightarrow β = \frac{4}{7}$

and $\frac{51}{14} = \frac{9 + γ}{2} \Rightarrow γ = \frac{- 12}{7}$

So, $14 (α + β + γ) = 14 \times \frac{54}{7} = 108$ .

Q 16 :

The distance of the point (7, –2, 11) from the line $\frac{x - 6}{1} = \frac{y - 4}{0} = \frac{z - 8}{3}$ along the line $\frac{x - 5}{2} = \frac{y - 1}{- 3} = \frac{z - 5}{6}$ is : [2024]

18
12
21
14

1

2

3

4

(4)

Let A be the point (7, –2, 11).

Equation of line AB is $\frac{x - 7}{2} = \frac{y + 2}{- 3} = \frac{z - 11}{6} = λ$

$\Rightarrow x = 2 λ + 7, y = - 3 λ - 2 and z = 6 λ + 11$

Point B is of the form $(2 λ + 7, - 3 λ - 2, 6 λ + 11)$ .

Now, point B lies on the line $\frac{x - 6}{1} = \frac{y - 4}{0} = \frac{z - 8}{3}$

$\Rightarrow \frac{2 λ + 7 - 6}{1} = \frac{- 3 λ - 2 - 4}{0} = \frac{6 λ + 11 - 8}{3}$

$\Rightarrow - 3 λ - 6 = 0 \Rightarrow λ = - 2$

Thus point B is (3, 4, –1).

$∴$ Required distance = $A B = \sqrt{4^{2} + {(- 6)}^{2} + 12^{2}}$

. $= \sqrt{16 + 36 + 144} = \sqrt{196} = 14 units$ .

Q 17 :

If the shortest distance between the lines $\frac{x - 4}{1} = \frac{y + 1}{2} = \frac{z}{- 3}$ and $\frac{x - λ}{2} = \frac{y + 1}{4} = \frac{z - 2}{- 5}$ is $\frac{6}{\sqrt{5}}$ , then the sum of all possible values of $λ$ is : [2024]

5
7
8
10

1

2

3

4

(3)

Here $x_{1} = 4, y_{1} = - 1, z_{1} = 0$ and $x_{2} = λ, y_{2} = - 1, z_{2} = 2$

$a_{1} = 1, b_{1} = 2, c_{1} = - 3$ and $a_{2} = 2, b_{2} = 4, c_{2} = - 5$

$∵ \frac{6}{\sqrt{5}} =$ $| \frac{| \begin{matrix} λ - 4 & 0 & 2 \\ 1 & 2 & - 3 \\ 2 & 4 & - 5 \end{matrix} |}{\sqrt{{(- 10 + 12)}^{2} + {(- 6 + 5)}^{2} + {(4 - 4)}^{2}}} |$

$\Rightarrow \frac{6}{\sqrt{5}} =$ $| \frac{2 λ - 8}{\sqrt{5}} |$ $\Rightarrow λ = 7 or 1$ .

Q 18 :

Let the image of the point (1, 0, 7) in the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ be the point $(α, β, γ)$ . Then which one of the following points lies on the line passing through $(α, β, γ)$ and making angles $\frac{2 π}{3}$ and $\frac{3 π}{4}$ with y-axis and z-axis respectively and an acute angle with x-axis? [2024]

$(3, - 4, 3 + 2 \sqrt{2})$
$(1, - 2, 1 + \sqrt{2})$
$(1, 2, 1 - \sqrt{2})$
$(3, 4, 3 - 2 \sqrt{2})$

1

2

3

4

(4)

Let $L_{1} : \frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} = λ$ (say)

$\Rightarrow x = λ, y = 2 λ + 1, z = 3 λ + 2$

Let the coordinates of M are $(λ, 2 λ + 1, 3 λ + 2)$ lie on the given line.

Let given point is P(1, 0, 7).

$∴$ Direction ratio of PM are $(λ - 1, 2 λ + 1, 3 λ - 5)$ PM is perpendicular the given line $L_{1}$ .

$∴ 1 (λ - 1) + 2 (2 λ + 1) + 3 (3 λ - 5) = 0$

$\Rightarrow λ - 1 + 4 λ + 2 + 9 λ - 15 = 0$

$\Rightarrow 14 λ - 14 = 0 \Rightarrow λ = 1$

$∴$ The coordinates of M are (1, 3, 5).

$∵$ M is mid point of PQ.

So, $\frac{α + 1}{2} = 1, \frac{β}{2} = 3$ and $\frac{γ + 7}{2} = 5$

$\Rightarrow α = 1, β = 6 and γ = 3$

$∴$ The image of point P(1, 0, 7) is Q(1, 6, 3).

Now the direction cosine of the line are

$m = \frac{\cos 2 π}{3} = \cos (π - \frac{π}{3}) = - \cos \frac{π}{3} = - \frac{1}{2}$

$n = \cos (\frac{3 π}{4}) = \cos (π - \frac{π}{4}) = - \frac{1}{\sqrt{2}}$

We know that, $l^{2} + m^{2} + n^{2} = 1$

$\Rightarrow l^{2} + \frac{1}{4} + \frac{1}{2} = 1$

$\Rightarrow l^{2} = 1 - \frac{1}{4} - \frac{1}{2} = \frac{4 - 1 - 2}{4} = \frac{1}{4} \Rightarrow l = \pm \frac{1}{2}$

$\Rightarrow l = \frac{1}{2}$ ( $∵$ Line make an acute angle with x-axis)

The equation of line passing through (1, 6, 3) with direction cosines $\frac{1}{2}, - \frac{1}{2}$ and $- \frac{1}{\sqrt{2}}$ is given by

$\vec{r} = \hat{i} + 6 \hat{j} + 3 \hat{k} + λ (\frac{1}{2} \hat{i} - \frac{1}{2} \hat{j} - \frac{1}{\sqrt{2}} \hat{k})$

$= (1 + \frac{λ}{2}) \hat{i} + (6 - \frac{λ}{2}) \hat{j} + (3 - \frac{λ}{\sqrt{2}}) \hat{k}$

Only option (4) satisfied the above equation for $λ$ = 4.

Q 19 :

Let PQR be a triangle with R(–1, 4, 2). Suppose M(2, 1, 2) is the mid point of PQ. The distance of the centroid of $△ P Q R$ from the point of intersection of the lines $\frac{x - 2}{0} = \frac{y}{2} = \frac{z + 3}{- 1}$ and $\frac{x - 1}{1} = \frac{y + 3}{- 3} = \frac{z + 1}{1}$ is [2024]

$\sqrt{69}$
69
$\sqrt{99}$
9

1

2

3

4

(1)

Let centroid G divides MR in the ratio 1 : 2.

So, centroid is given by

$∴ G (\frac{4 - 1}{3}, \frac{2 + 4}{3}, \frac{4 + 2}{3})$ i.e., G(1, 2, 2)

Let $l_{1} : \frac{x - 2}{0} = \frac{y}{2} = \frac{z + 3}{- 1} = λ$ (say)

$\Rightarrow x = 2, y = 2 λ, z = - λ - 3$

$l_{2} : \frac{x - 1}{1} = \frac{y + 3}{- 3} = \frac{z + 1}{1} = r$ (say)

$\Rightarrow$ x = r + 1, y = –3r – 3, z = r – 1

Now, $2 = r + 1 \Rightarrow r = 1$ and $2 λ = - 3 r - 3 \Rightarrow λ = - 3$ .

$∴$ Point of intersection of lines $l_{1}$ and $l_{2}$ is given by A(2, –6, 0).

$∴$ Required distance, $A G = \sqrt{1 + 64 + 4} = \sqrt{69}$ .

Q 20 :

Let P(3, 2, 3), Q(4, 6, 2) and R(7, 3, 2) be the vertices of $△ P Q R$ . Then the angle $∠ Q P R$ is [2024]

$\cos^{- 1} (\frac{7}{18})$
$\cos^{- 1} (\frac{1}{18})$
$\frac{π}{3}$
$\frac{π}{6}$

1

2

3

4

(3)

Direction ratios of line PQ = <4 – 3, 6 – 2, 2 – 3> = <1, 4, –1>

Direction ratios of line PR = <7 – 3, 3 – 2, 2 – 3> = <4, 1, –1>

Let $θ$ be the required angle.

$∴ \cos θ =$ $| \frac{1 \times 4 + 4 \times 1 + (- 1) \times (- 1)}{\sqrt{1 + 16 + 1} \sqrt{16 + 1 + 1}} |$ $=$ $| \frac{9}{18} |$ $= \frac{1}{2}$

$\Rightarrow θ = \cos^{- 1} (\frac{1}{2}) = \frac{π}{3} ∴ ∠ Q P R = \frac{π}{3}$ .

Q 21 :

Let $(α, β, γ)$ be the foot of the perpendicular from the point (1, 2, 3) on the line $\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3}$ . Then $19 (α + β + γ)$ is equal to [2024]

100
99
101
102

1

2

3

4

(3)

Let $\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3} = λ$

$\Rightarrow (α, β, γ) = (5 λ - 3, 2 λ + 1, 3 λ - 4)$

Let $B (α, β, γ)$ be the foot of perpendicular from A on the given line

Now, $5 \hat{i} + 2 \hat{j} + 3 \hat{k}$ is the direction vector of given line.

$\Rightarrow \vec{A B} \cdot (5 \hat{i} + 2 \hat{j} + 3 \hat{k}) = 0$

$\Rightarrow (5 λ - 4) 5 + (2 λ - 1) 2 + (3 λ - 7) 3 = 0$

$\Rightarrow 25 λ + 4 λ + 9 λ = 20 + 2 + 21$

$\Rightarrow 38 λ = 43 \Rightarrow λ = \frac{43}{38}$

$∴ 19 (α + β + γ) = 19 (5 λ - 3 + 2 λ + 1 + 3 λ - 4) = 19 (10 λ - 6)$

$= 19 (10 \times \frac{43}{38} - 6) = 215 - 114 = 101$ .

Q 22 :

Let $L_{1} : \vec{r} = (\hat{i} - \hat{j} + 2 \hat{k}) + λ (\hat{i} - \hat{j} + 2 \hat{k}), λ \in R$

$L_{2} : \vec{r} = (\hat{j} - \hat{k}) + μ (3 \hat{i} + \hat{j} + p \hat{k}), μ \in R$

$L_{3} : \vec{r} = δ (l \hat{i} + m \hat{j} + n \hat{k}), δ \in R$

be three lines such that $L_{1}$ is perpendicular to $L_{2}$ and $L_{3}$ is perpendicular to both $L_{1}$ and $L_{2}$ . Then, the point which lies on $L_{3}$ is [2024]

(1, –7, 4)
(–1, 7, 4)
(–1, –7, 4)
(1, 7, –4)

1

2

3

4

(2)

Given, $L_{1} : \vec{r} = (\hat{i} - \hat{j} + 2 \hat{k}) + λ (\hat{i} - \hat{j} + 2 \hat{k}), λ \in R$

$L_{2} : \vec{r} = (\hat{j} - \hat{k}) + μ (3 \hat{i} + \hat{j} + p \hat{k}), μ \in R$

and $L_{3} : \vec{r} = δ (l \hat{i} + m \hat{j} + n \hat{k}), δ \in R$

since, $L_{1}$ is perpendicular to $L_{2}$ .

$∴ (\hat{i} - \hat{j} + 2 \hat{k}) \cdot (3 \hat{i} + \hat{j} + p \hat{k}) = 0$

$\Rightarrow 3 - 1 + 2 p = 0 \Rightarrow p = - 1$

Also, $L_{3}$ is perpendicular to both $L_{1}$ and $L_{2}$ .

$∴ L_{3} is parallel to (\hat{i} - \hat{j} + 2 \hat{k}) \times (3 \hat{i} + \hat{j} + p \hat{k})$

Now, $(\hat{i} - \hat{j} + 2 \hat{k}) \times (3 \hat{i} + \hat{j} + p \hat{k})$

$=$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 1 & 2 \\ 3 & 1 & p \end{matrix} |$ $= \hat{i} (- p - 2) - \hat{j} (p - 6) + \hat{k} (1 + 3)$

$= - \hat{i} + 7 \hat{j} + 4 \hat{k}$ $(∵ p = - 1)$

$∴ l \hat{i} + m \hat{j} + n \hat{k} = - \hat{i} + 7 \hat{j} + 4 \hat{k}$

$\Rightarrow l = - 1, m = 7 and n = 4$

$∴ (- 1, 7, 4) lies on L_{3}$ .

Q 23 :

The distance of the point Q(0, 2, –2) from the line passing through the point P(5, –4, 3) and perpendicular to the lines $\vec{r} = (- 3 \hat{i} + 2 \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 5 \hat{k}), λ \in R$ and $\vec{r} = (\hat{i} - 2 \hat{j} + \hat{k}) + μ (- \hat{i} + 3 \hat{j} + 2 \hat{k}), μ \in R$ is : [2024]

$\sqrt{74}$
$\sqrt{54}$
$\sqrt{86}$
$\sqrt{20}$

1

2

3

4

(1)

Given, $\vec{r} = (- 3 \hat{i} + 2 \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 5 \hat{k})$

and $\vec{r} = (\hat{i} - 2 \hat{j} + \hat{k}) + μ (- \hat{i} + 3 \hat{j} + 2 \hat{k})$

${\vec{b}}_{1} \times {\vec{b}}_{2} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 5 \\ - 1 & 3 & 2 \end{matrix} |$

${\vec{b}}_{1} \times {\vec{b}}_{2} = \hat{i} (6 - 15) - \hat{j} (4 + 5) + \hat{k} (6 + 3)$

${\vec{b}}_{1} \times {\vec{b}}_{2} = - 9 \hat{i} - 9 \hat{j} + 9 \hat{k}$

Then the equation of the line

$l = \frac{x - 5}{1} = \frac{y + 4}{1} = \frac{z - 3}{- 1} = λ$

$x = λ + 5; y = λ - 4; z = 3 - λ$

$∴ \vec{Q R} \cdot l = 0$

$(λ + 5) \cdot 1 + (λ - 6) 1 + (3 - λ + 2) (- 1) = 0$

$λ + 5 + λ - 6 + (- 5) + λ = 0$

$λ + 5 + λ - 6 - 5 + λ = 0$

$3 λ - 6 = 0$

$λ = \frac{6}{3}$

$λ = 2$

Then, R = (7, –2, 1)

$Q R = \sqrt{49 + 16 + 9} \Rightarrow Q R = \sqrt{74}$ .

Q 24 :

Let $(α, β, γ)$ be the mirror image of the point (2, 3, 5) in the line $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ . Then $2 α + 3 β + 4 γ$ is equal to [2024]

34
32
31
33

1

2

3

4

(4)

Given equation of the line is $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$

Let A(2, 3, 5) be the given point and let B be the foot of the perpendicular drawn from the point A on the line

$\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} = λ, λ \in R$

$\Rightarrow x = 2 λ + 1, y = 3 λ + 2, z = 4 λ + 3$

$∴ B \equiv (2 λ + 1, 3 λ + 2, 4 λ + 3)$

Direction ratios of AB are $2 λ - 1, 3 λ - 1, 4 λ - 2$ .

Now, AB is perpendicular to the given line.

$∴ 2 (2 λ - 1) + 3 (3 λ - 1) + 4 (4 λ - 2) = 0$

$\Rightarrow 4 λ - 2 + 9 λ - 3 + 16 λ - 8 = 0$

$\Rightarrow 29 λ - 13 = 0 \Rightarrow λ = \frac{13}{29}$

$∴$ Coordinates of B are $(\frac{55}{29}, \frac{97}{29}, \frac{139}{29})$

Now, using mid point formula,

$\frac{55}{29} = \frac{2 + α}{2}; \frac{97}{29} = \frac{3 + β}{2}; \frac{139}{29} = \frac{5 + γ}{2}$

$\Rightarrow α = \frac{52}{29}, β = \frac{107}{29}, γ = \frac{133}{29}$

$∴ 2 α + 3 β + 4 γ = 2 \times \frac{52}{29} + \frac{3 \times 107}{29} + \frac{4 \times 133}{29} = \frac{957}{29} = 33$ .

Q 25 :

The shortest distance, between lines $L_{1}$ and $L_{2}$ , where $L_{1} : \frac{x - 1}{2} = \frac{y + 1}{- 3} = \frac{z + 4}{2}$ and $L_{2}$ is the line, passing through the points A(–4, 4, 3), B(–1, 6, 3) and perpendicular to the line $\frac{x - 3}{- 2} = \frac{y}{3} = \frac{z - 1}{1}$ , is [2024]

$\frac{141}{\sqrt{221}}$
$\frac{42}{\sqrt{117}}$
$\frac{121}{\sqrt{221}}$
$\frac{24}{\sqrt{117}}$

1

2

3

4

(1)

We have, $L_{1} = \frac{x - 1}{2} = \frac{y + 1}{- 3} = \frac{z + 4}{2}$

Equation of the line passing through the points A(–4, 4, 3) and B(–1, 6, 3) is

$L_{2} = \frac{x + 4}{3} = \frac{y - 4}{2} = \frac{z - 3}{0}$

Here, $a_{1} = 2, b_{1} = - 3, c_{1} = 2; x_{1} = 1, y_{1} = - 1, z_{1} = - 4$

$a_{2} = 3, b_{2} = 2, c_{2} = 0; x_{2} = - 4, y_{2} = 4, z_{2} = 3$

$∴$ Required shortest distance

$d =$ $| \frac{| \begin{matrix} - 5 & 5 & 7 \\ 2 & - 3 & 2 \\ 3 & 2 & 0 \end{matrix} |}{\sqrt{{(4 + 9)}^{2} + {(0 - 4)}^{2} + {(6 - 0)}^{2}}} |$

$=$ $| \frac{- 5 (0 - 4) - 5 (0 - 6) + 7 (4 + 9)}{\sqrt{169 + 16 + 36}} |$ $=$ $| \frac{20 + 30 + 91}{\sqrt{221}} |$ $= \frac{141}{\sqrt{221}}$ .

Q 26 :

If the shortest distance between the lines $\frac{x + 2}{2} = \frac{y + 3}{3} = \frac{z - 5}{4}$ and $\frac{x - 3}{1} = \frac{y - 2}{- 3} = \frac{z + 4}{2}$ is $\frac{38}{3 \sqrt{5}} k$ , and $\int_{0}^{k} [x^{2}] d x = α - \sqrt{α}$ , where [x] denotes the greatest integer function, then $6 α^{3}$ is equal to __________. [2024]

(48)

$L_{1} : \frac{x + 2}{2} = \frac{y + 3}{3} = \frac{z - 5}{4}$

$L_{2} : \frac{x - 3}{1} = \frac{y - 2}{- 3} = \frac{z + 4}{2}$

$∴ {\vec{a}}_{1} = - 2 \hat{i} - 3 \hat{j} + 5 \hat{k}, {\vec{a}}_{2} = 3 \hat{i} + 2 \hat{j} - 4 \hat{k}$

${\vec{b}}_{1} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}, {\vec{b}}_{2} = \hat{i} - 3 \hat{j} + 2 \hat{k}$

Now, ${\vec{a}}_{2} - {\vec{a}}_{1} = 5 \hat{i} + 5 \hat{j} - 9 \hat{k}$

Shortest distance between the lines = $| \frac{({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2})}{| {\vec{b}}_{1} \times {\vec{b}}_{2} |} |$ and $| {\vec{b}}_{1} \times {\vec{b}}_{2} |$ $=$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 1 & - 3 & 2 \end{matrix} |$

$= \hat{i} (6 + 12) - \hat{j} (4 - 4) + \hat{k} (- 6 - 3) = 18 \hat{i} - 9 \hat{k}$

$d =$ $| \frac{(5 \hat{i} + 5 \hat{j} - 9 \hat{k}) \cdot (18 \hat{i} - 9 \hat{k})}{\sqrt{324 + 81}} |$ $=$ $| \frac{90 + 81}{9 \sqrt{5}} |$ $= \frac{171}{9 \sqrt{5}}$

From the question, we get $\frac{38}{3 \sqrt{5}} k = \frac{171}{9 \sqrt{5}} \Rightarrow k = \frac{3}{2}$

$∴ \int_{0}^{k} [x^{2}] d x = \int_{0}^{\frac{3}{2}} [x^{2}] d x = \int_{0}^{1} 0 d x + \int_{1}^{\sqrt{2}} 1 d x + \int_{\sqrt{2}}^{\frac{3}{2}} 2 d x$

$= 0 + (\sqrt{2} - 1) + 2 (\frac{3}{2} - \sqrt{2}) = \sqrt{2} - 1 + 3 - 2 \sqrt{2} = 2 - \sqrt{2}$

$= α - \sqrt{α}$ (Given)

Hence, $α = 2$

$∴ 6 α^{3} = 6 \times 8 = 48$ .

Q 27 :

Consider a line L passing through the points P(1, 2, 1) and Q(2, 1, –1). If the mirror image of the point A(2, 2, 2) in the line L is $(α, β, γ)$ , then $α + β + 6 γ$ is equal to __________. [2024]

(6)

PQ : $\frac{x - 1}{1} = \frac{y - 2}{- 1} = \frac{z - 1}{- 2} = λ$ (say)

Any point on the line PQ is of the form

$B (λ + 1, - λ + 2, - 2 λ + 1)$

$\vec{A B} = (λ + 1 - 2) \hat{i} + (- λ + 2 - 2) \hat{j} + (- 2 λ + 1 - 2) \hat{k}$

$\Rightarrow \vec{A B} = (λ - 1) \hat{i} - λ \hat{j} + (- 2 λ - 1) \hat{k} and \vec{P Q} = 1 \hat{i} - \hat{j} - 2 \hat{k}$

$\vec{A B} \cdot \vec{P Q} = 0$

$\Rightarrow (λ - 1) + λ + 2 (2 λ + 1) = 0$

$\Rightarrow λ - 1 + λ + 4 λ + 2 = 0$

$\Rightarrow 6 λ = - 1 \Rightarrow λ = \frac{- 1}{6}$

$∴ B (\frac{- 1}{6} + 1, \frac{1}{6} + 2, - 2 (\frac{- 1}{6}) + 1) = B (\frac{5}{6}, \frac{13}{6}, \frac{8}{6})$

Now B is mid point of AA'

$\Rightarrow (\frac{5}{6}, \frac{13}{6}, \frac{8}{6}) = (\frac{α + 2}{2}, \frac{β + 2}{2}, \frac{γ + 2}{2})$

$\Rightarrow α + 2 = \frac{10}{6}, β + 2 = \frac{26}{6}, γ + 2 = \frac{16}{6}$

$\Rightarrow α = \frac{- 1}{3}, β = \frac{7}{3}, γ = \frac{2}{3}$

Hence, $α + β + 6 γ = \frac{- 1}{3} + \frac{7}{3} + 6 \times \frac{2}{3} = \frac{18}{3} = 6$ .

Q 28 :

Let the point $(- 1, α, β)$ lie on the line of the shortest distance between the lines $\frac{x + 2}{- 3} = \frac{y - 2}{4} = \frac{z - 5}{2}$ and $\frac{x + 2}{- 1} = \frac{y + 6}{2} = \frac{z - 1}{0}$ . Then ${(α - β)}^{2}$ is equal to _________. [2024]

(25)

Let $L_{1} : \frac{x + 2}{- 3} = \frac{y - 2}{4} = \frac{z - 5}{2}$ and $L_{2} : \frac{x + 2}{- 1} = \frac{y + 6}{2} = \frac{z - 1}{0}$

Let $P (- 3 λ - 2, 4 λ + 2, 2 λ + 5)$ be point on $L_{1}$ and $Q (- μ - 2, 2 μ - 6, 1)$ be point on $L_{2}$ .

Direction ratios of PQ = $(3 λ - μ, 2 μ - 4 λ - 8, - 2 λ - 4)$

Also, $\vec{P Q} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - 3 & 4 & 2 \\ - 1 & 2 & 0 \end{matrix} |$ $= - 4 \hat{i} - 2 \hat{j} - 2 \hat{k}$

i.e., $2 \hat{i} + \hat{j} + \hat{k}$

$\Rightarrow \frac{3 λ - μ}{2} = \frac{2 μ - 4 λ - 8}{1} = \frac{- 2 λ - 4}{1}$

$\Rightarrow 3 λ - μ = - 4 λ - 8 and 2 μ - 4 λ - 8 = - 2 λ - 4$

$\Rightarrow μ = 7 λ + 8 and μ = λ + 2$

$\Rightarrow λ = - 1 and μ = 1$

$∴$ Equation of PQ is, $\frac{x + 3}{2} = \frac{y + 4}{1} = \frac{z - 1}{1}$

Now, $(- 1, α, β)$ lies on PQ

$\Rightarrow 1 = α + 4 = β - 1$

$\Rightarrow α = - 3, β = 2$

$∴ {(α - β)}^{2} = 25$ .

Q 29 :

Let P be the point (10, –2, –1) and Q be the foot of the perpendicular drawn from the point R(1, 7, 6) on the line passing through the points (2, –5, 11) and (–6, 7, –5). Then the length of the line segment PQ is equal to __________. [2024]

(13)

Equation of line passing through (2, –5, 11) and (–6, 7, –5) is

$L : \frac{x + 6}{8} = \frac{y - 7}{- 12} = \frac{z + 5}{16}$

i.e., $\frac{x + 6}{2} = \frac{y - 7}{- 3} = \frac{z + 5}{4} = λ$ ... (i)

$\Rightarrow$ any point Q on the given line is $(2 λ - 6, - 3 λ + 7, 4 λ - 5)$

Now, D.r.'s of QR = $(2 λ - 7, - 3 λ, 4 λ - 11)$

Since QR is perpendicular to L

$\Rightarrow 2 (2 λ - 7) - 3 (- 3 λ) + 4 (4 λ - 11) = 0$

$\Rightarrow 29 λ - 58 = 0 \Rightarrow λ = 2$

So Q(–2, 1, 3) is the required point

and $| \vec{P Q} |$ $= \sqrt{{(- 2 - 10)}^{2} + {(1 + 2)}^{2} + {(3 + 1)}^{2}}$ = $\sqrt{169} = 13$ .

Q 30 :

If the shortest distance between the lines $\frac{x - λ}{3} = \frac{y - 2}{- 1} = \frac{z - 1}{1}$ and $\frac{x + 2}{- 3} = \frac{y + 5}{2} = \frac{z - 4}{4}$ is $\frac{44}{\sqrt{30}}$ , then the largest possible value of $| λ |$ is equal to __________. [2024]

(43)

We have, $\vec{a_{1}} = λ \hat{i} + 2 \hat{j} + \hat{k}, \vec{a_{2}} = - 2 \hat{i} - 5 \hat{j} + 4 \hat{k}$

$\vec{n_{1}} = 3 \hat{i} - \hat{j} + \hat{k}, \vec{n_{2}} = - 3 \hat{i} + 2 \hat{j} + 4 \hat{k}$

Now, $\vec{n_{1}} \times \vec{n_{2}} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & - 1 & 1 \\ - 3 & 2 & 4 \end{matrix} |$ $= - 6 \hat{i} - 15 \hat{j} + 3 \hat{k}$

Shortest distance between lines = $| \frac{(\vec{a_{2}} - \vec{a_{1}}) \cdot (\vec{n_{1}} \times \vec{n_{2}})}{| \vec{n_{1}} \times \vec{n_{2}} |} |$

$\Rightarrow \frac{44}{\sqrt{30}} =$ $| \frac{(- 2 - λ) \vec{i} - 7 \vec{j} + 3 \vec{k}) \cdot (- 6 \vec{i} - 15 \vec{j} + 3 \vec{k})}{\sqrt{36 + 225 + 9}} |$

$\Rightarrow \frac{44}{\sqrt{30}} =$ $| \frac{- 6 (- 2 - λ) + 105 + 9}{\sqrt{270}} |$ $\Rightarrow$ $| \frac{6 λ + 126}{\sqrt{270}} |$ $= \frac{44}{\sqrt{30}}$

$\Rightarrow | 6 λ + 126 | = 132 \Rightarrow | λ + 21 | = 22$

$\Rightarrow λ + 21 = \pm 22 \Rightarrow | λ |_{\max} = 43$ .

Topic Question Set

Q 1 :

Let the point, on the line passing through the points P(1, –2, 3) and Q(5, –4, 7), farther from the origin and at a distance of 9 units from the point P, be $(α, β, γ)$ . Then $α^{2} + β^{2} + γ^{2}$ is equal to [2024]

Q 2 :

Let P be the point of intersection of the lines $\frac{x - 2}{1} = \frac{y - 4}{5} = \frac{z - 2}{1}$ and $\frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 3}{2}$ . Then, the shortest distance of P from the line 4x = 2y = z is [2024]

Q 3 :

Let d be the distance of the point of intersection of the lines $\frac{x + 6}{3} = \frac{y}{2} = \frac{z + 1}{1}$ and $\frac{x - 7}{4} = \frac{y - 9}{3} = \frac{z - 4}{2}$ from the point(7, 8, 9). Then $d^{2} + 6$ is equal to [2024]

Q 4 :

If the line $\frac{2 - x}{3} = \frac{3 y - 2}{4 λ + 1} = 4 - z$ makes a right angle with the line $\frac{x + 3}{3 μ} = \frac{1 - 2 y}{6} = \frac{5 - z}{7}$ , then $4 λ + 9 μ$ is equal to : [2024]

Q 5 :

Let $(α, β, γ)$ be the image of the point (8, 5, 7) in the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 2}{5}$ . Then $α + β + γ$ is equal to : [2024]

Q 6 :

The shortest distance between the lines $\frac{x - 3}{2} = \frac{y + 15}{- 7} = \frac{z - 9}{5}$ and $\frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{- 3}$ is [2024]

Q 7 :

Let $P (α, β, γ)$ be the image of the point Q(3, –3, 1) in the line $\frac{x - 0}{1} = \frac{y - 3}{1} = \frac{z - 1}{- 1}$ and R be the point (2, 5, –1). If the area of the triangle PQR is $λ$ and $λ^{2} = 14 K$ , then K is equal to : [2024]

Q 9 :

If the shortest distance between the lines $\frac{x - λ}{2} = \frac{y - 4}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{4} = \frac{y - 4}{6} = \frac{z - 7}{8}$ is $\frac{13}{\sqrt{29}}$ , then a value of $λ$ is [2024]

Q 10 :

The shortest distance between the lines $\frac{x - 3}{4} = \frac{y + 7}{- 11} = \frac{z - 1}{5}$ and $\frac{x - 5}{3} = \frac{y - 9}{- 6} = \frac{z + 2}{1}$ is [2024]

Q 11 :

Let the line L intersect the lines x – 2 = – y = z – 1, 2(x + 1) = 2(y –1) = z + 1 and be parallel to the line $\frac{x - 2}{3} = \frac{y - 1}{1} = \frac{z - 2}{2}$ . Then which of the following points lis on L? [2024]

Q 12 :

Consider the line L passing through the points (1, 2, 3) and (2, 3, 5). The distance of the point $(\frac{11}{3}, \frac{11}{3}, \frac{19}{3})$ from the line L along the line $\frac{3 x - 11}{2} = \frac{3 y - 11}{1} = \frac{3 z - 19}{2}$ is equal to [2024]

Q 13 :

If the shortest distance between the lines $\frac{x - λ}{- 2} = \frac{y - 2}{1} = \frac{z - 1}{1}$ and $\frac{x - \sqrt{3}}{1} = \frac{y - 1}{- 2} = \frac{z - 2}{1}$ is 1, then the sum of all possible values of $λ$ is: [2024]

Q 14 :

Let P and Q be the points on the line $\frac{x + 3}{8} = \frac{y - 4}{2} = \frac{z + 1}{2}$ which are at a distance of 6 units from the point R(1, 2, 3). If the centroid of the triangle PQR is $(α, β, γ)$ , then $α^{2} + β^{2} + γ^{2}$ is : [2024]

Q 15 :

If the mirror image of the point P(3, 4, 9) in the line $\frac{x - 1}{3} = \frac{y + 1}{2} = \frac{z - 2}{1}$ is $(α, β, γ)$ , then $14 (α + β + γ)$ is : [2024]

Q 16 :

The distance of the point (7, –2, 11) from the line $\frac{x - 6}{1} = \frac{y - 4}{0} = \frac{z - 8}{3}$ along the line $\frac{x - 5}{2} = \frac{y - 1}{- 3} = \frac{z - 5}{6}$ is : [2024]

Q 17 :

If the shortest distance between the lines $\frac{x - 4}{1} = \frac{y + 1}{2} = \frac{z}{- 3}$ and $\frac{x - λ}{2} = \frac{y + 1}{4} = \frac{z - 2}{- 5}$ is $\frac{6}{\sqrt{5}}$ , then the sum of all possible values of $λ$ is : [2024]

Q 20 :

Let P(3, 2, 3), Q(4, 6, 2) and R(7, 3, 2) be the vertices of $△ P Q R$ . Then the angle $∠ Q P R$ is [2024]

Q 21 :

Let $(α, β, γ)$ be the foot of the perpendicular from the point (1, 2, 3) on the line $\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3}$ . Then $19 (α + β + γ)$ is equal to [2024]

Q 24 :

Let $(α, β, γ)$ be the mirror image of the point (2, 3, 5) in the line $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ . Then $2 α + 3 β + 4 γ$ is equal to [2024]

Q 27 :

Consider a line L passing through the points P(1, 2, 1) and Q(2, 1, –1). If the mirror image of the point A(2, 2, 2) in the line L is $(α, β, γ)$ , then $α + β + 6 γ$ is equal to __________. [2024]

Q 28 :

Let the point $(- 1, α, β)$ lie on the line of the shortest distance between the lines $\frac{x + 2}{- 3} = \frac{y - 2}{4} = \frac{z - 5}{2}$ and $\frac{x + 2}{- 1} = \frac{y + 6}{2} = \frac{z - 1}{0}$ . Then ${(α - β)}^{2}$ is equal to _________. [2024]

Q 29 :

Let P be the point (10, –2, –1) and Q be the foot of the perpendicular drawn from the point R(1, 7, 6) on the line passing through the points (2, –5, 11) and (–6, 7, –5). Then the length of the line segment PQ is equal to __________. [2024]

Q 30 :

If the shortest distance between the lines $\frac{x - λ}{3} = \frac{y - 2}{- 1} = \frac{z - 1}{1}$ and $\frac{x + 2}{- 3} = \frac{y + 5}{2} = \frac{z - 4}{4}$ is $\frac{44}{\sqrt{30}}$ , then the largest possible value of $| λ |$ is equal to __________. [2024]

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Topic Question Set

Q 1 : Let the point, on the line passing through the points P(1, –2, 3) and Q(5, –4, 7), farther from the origin and at a distance of 9 units from the point P, be (α,β,γ). Then α2+β2+γ2 is equal to [2024]

Q 2 : Let P be the point of intersection of the lines x–21=y–45=z–21 and x–32=y–23=z–32. Then, the shortest distance of P from the line 4x = 2y = z is [2024]

Q 3 : Let d be the distance of the point of intersection of the lines x+63=y2=z+11 and x–74=y–93=z–42 from the point(7, 8, 9). Then d2+6 is equal to [2024]

Q 4 : If the line 2–x3=3y–24λ+1=4–z makes a right angle with the line x+33μ=1–2y6=5–z7, then 4λ+9μ is equal to : [2024]

Q 5 : Let (α, β, γ) be the image of the point (8, 5, 7) in the line x–12=y+13=z–25. Then α+β+γ is equal to : [2024]

Q 6 : The shortest distance between the lines x–32=y+15–7=z–95 and x+12=y–11=z–9–3 is [2024]

Q 7 : Let P(α, β, γ) be the image of the point Q(3, –3, 1) in the line x–01=y–31=z–1–1 and R be the point (2, 5, –1). If the area of the triangle PQR is λ and λ2=14K, then K is equal to : [2024]

Q 8 : If the shortest distance between the lines L1:r→=(2+λ)i^+(1–3λ)j^+(3+4λ)k^, λ∈R L2:r→=2(1+μ)i^+3(1+μ)j^+(5+μ)k^, μ∈R is mn, where gcd(m, n) = 1, then the value of m + n equals [2024]

Q 9 : If the shortest distance between the lines x–λ2=y–43=z–34 and x–24=y–46=z–78 is 1329, then a value of λ is [2024]

Q 10 : The shortest distance between the lines x–34=y+7–11=z–15 and x–53=y–9–6=z+21 is [2024]

Q 11 : Let the line L intersect the lines x – 2 = – y = z – 1, 2(x + 1) = 2(y –1) = z + 1 and be parallel to the line x–23=y–11=z–22. Then which of the following points lis on L? [2024]

Q 12 : Consider the line L passing through the points (1, 2, 3) and (2, 3, 5). The distance of the point (113,113,193) from the line L along the line 3x–112=3y–111=3z–192 is equal to [2024]

Q 13 : If the shortest distance between the lines x–λ–2=y–21=z–11 and x–31=y–1–2=z–21 is 1, then the sum of all possible values of λ is: [2024]

Q 14 : Let P and Q be the points on the line x+38=y–42=z+12 which are at a distance of 6 units from the point R(1, 2, 3). If the centroid of the triangle PQR is (α,β,γ), then α2+β2+γ2 is : [2024]

Q 15 : If the mirror image of the point P(3, 4, 9) in the line x-13=y+12=z-21 is (α,β,γ), then 14(α+β+γ) is : [2024]

Q 16 : The distance of the point (7, –2, 11) from the line x–61=y–40=z–83 along the line x–52=y–1–3=z–56 is : [2024]

Q 17 : If the shortest distance between the lines x–41=y+12=z–3 and x–λ2=y+14=z–2–5 is 65, then the sum of all possible values of λ is : [2024]

Q 18 : Let the image of the point (1, 0, 7) in the line x1=y–12=z–23 be the point (α, β, γ). Then which one of the following points lies on the line passing through (α, β, γ) and making angles 2π3 and 3π4 with y-axis and z-axis respectively and an acute angle with x-axis? [2024]

Q 19 : Let PQR be a triangle with R(–1, 4, 2). Suppose M(2, 1, 2) is the mid point of PQ. The distance of the centroid of △PQR from the point of intersection of the lines x–20=y2=z+3–1 and x–11=y+3–3=z+11 is [2024]

Q 20 : Let P(3, 2, 3), Q(4, 6, 2) and R(7, 3, 2) be the vertices of △PQR. Then the angle ∠QPR is [2024]

Q 21 : Let (α, β, γ) be the foot of the perpendicular from the point (1, 2, 3) on the line x+35=y–12=z+43. Then 19(α+β+γ) is equal to [2024]

Q 22 : Let L1:r→=(i^–j^+2k^)+λ(i^–j^+2k^), λ∈R L2:r→=(j^–k^)+μ(3i^+j^+pk^), μ∈R L3:r→=δ(li^+mj^+nk^), δ∈R be three lines such that L1 is perpendicular to L2 and L3 is perpendicular to both L1 and L2. Then, the point which lies on L3 is [2024]

Q 23 : The distance of the point Q(0, 2, –2) from the line passing through the point P(5, –4, 3) and perpendicular to the lines r→=(–3i^+2k^)+λ(2i^+3j^+5k^), λ∈R and r→=(i^–2j^+k^)+μ(–i^+3j^+2k^), μ∈R is : [2024]

Q 24 : Let (α, β, γ) be the mirror image of the point (2, 3, 5) in the line x–12=y–23=z–34. Then 2α+3β+4γ is equal to [2024]

Q 25 : The shortest distance, between lines L1 and L2, where L1:x–12=y+1–3=z+42 and L2 is the line, passing through the points A(–4, 4, 3), B(–1, 6, 3) and perpendicular to the line x–3–2=y3=z–11, is [2024]

Q 26 : If the shortest distance between the lines x+22=y+33=z–54 and x–31=y–2–3=z+42 is 3835k, and ∫0k[x2]dx=α–α, where [x] denotes the greatest integer function, then 6α3 is equal to __________. [2024]

Q 27 : Consider a line L passing through the points P(1, 2, 1) and Q(2, 1, –1). If the mirror image of the point A(2, 2, 2) in the line L is (α, β, γ), then α+β+6γ is equal to __________. [2024]

Q 28 : Let the point (–1, α, β) lie on the line of the shortest distance between the lines x+2–3=y–24=z–52 and x+2–1=y+62=z–10. Then (α–β)2 is equal to _________. [2024]

Q 29 : Let P be the point (10, –2, –1) and Q be the foot of the perpendicular drawn from the point R(1, 7, 6) on the line passing through the points (2, –5, 11) and (–6, 7, –5). Then the length of the line segment PQ is equal to __________. [2024]

Q 30 : If the shortest distance between the lines x–λ3=y–2–1=z–11 and x+2–3=y+52=z–44 is 4430, then the largest possible value of |λ| is equal to __________. [2024]

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Q 1 :

Let the point, on the line passing through the points P(1, –2, 3) and Q(5, –4, 7), farther from the origin and at a distance of 9 units from the point P, be $(α, β, γ)$ . Then $α^{2} + β^{2} + γ^{2}$ is equal to [2024]

Q 2 :

Let P be the point of intersection of the lines $\frac{x - 2}{1} = \frac{y - 4}{5} = \frac{z - 2}{1}$ and $\frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 3}{2}$ . Then, the shortest distance of P from the line 4x = 2y = z is [2024]

Q 3 :

Let d be the distance of the point of intersection of the lines $\frac{x + 6}{3} = \frac{y}{2} = \frac{z + 1}{1}$ and $\frac{x - 7}{4} = \frac{y - 9}{3} = \frac{z - 4}{2}$ from the point(7, 8, 9). Then $d^{2} + 6$ is equal to [2024]

Q 4 :

If the line $\frac{2 - x}{3} = \frac{3 y - 2}{4 λ + 1} = 4 - z$ makes a right angle with the line $\frac{x + 3}{3 μ} = \frac{1 - 2 y}{6} = \frac{5 - z}{7}$ , then $4 λ + 9 μ$ is equal to : [2024]

Q 5 :

Let $(α, β, γ)$ be the image of the point (8, 5, 7) in the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 2}{5}$ . Then $α + β + γ$ is equal to : [2024]

Q 6 :

The shortest distance between the lines $\frac{x - 3}{2} = \frac{y + 15}{- 7} = \frac{z - 9}{5}$ and $\frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{- 3}$ is [2024]

Q 7 :

Let $P (α, β, γ)$ be the image of the point Q(3, –3, 1) in the line $\frac{x - 0}{1} = \frac{y - 3}{1} = \frac{z - 1}{- 1}$ and R be the point (2, 5, –1). If the area of the triangle PQR is $λ$ and $λ^{2} = 14 K$ , then K is equal to : [2024]

Q 9 :

If the shortest distance between the lines $\frac{x - λ}{2} = \frac{y - 4}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{4} = \frac{y - 4}{6} = \frac{z - 7}{8}$ is $\frac{13}{\sqrt{29}}$ , then a value of $λ$ is [2024]

Q 10 :

The shortest distance between the lines $\frac{x - 3}{4} = \frac{y + 7}{- 11} = \frac{z - 1}{5}$ and $\frac{x - 5}{3} = \frac{y - 9}{- 6} = \frac{z + 2}{1}$ is [2024]

Q 11 :

Let the line L intersect the lines x – 2 = – y = z – 1, 2(x + 1) = 2(y –1) = z + 1 and be parallel to the line $\frac{x - 2}{3} = \frac{y - 1}{1} = \frac{z - 2}{2}$ . Then which of the following points lis on L? [2024]

Q 12 :

Consider the line L passing through the points (1, 2, 3) and (2, 3, 5). The distance of the point $(\frac{11}{3}, \frac{11}{3}, \frac{19}{3})$ from the line L along the line $\frac{3 x - 11}{2} = \frac{3 y - 11}{1} = \frac{3 z - 19}{2}$ is equal to [2024]

Q 13 :

If the shortest distance between the lines $\frac{x - λ}{- 2} = \frac{y - 2}{1} = \frac{z - 1}{1}$ and $\frac{x - \sqrt{3}}{1} = \frac{y - 1}{- 2} = \frac{z - 2}{1}$ is 1, then the sum of all possible values of $λ$ is: [2024]

Q 14 :

Let P and Q be the points on the line $\frac{x + 3}{8} = \frac{y - 4}{2} = \frac{z + 1}{2}$ which are at a distance of 6 units from the point R(1, 2, 3). If the centroid of the triangle PQR is $(α, β, γ)$ , then $α^{2} + β^{2} + γ^{2}$ is : [2024]

Q 15 :

If the mirror image of the point P(3, 4, 9) in the line $\frac{x - 1}{3} = \frac{y + 1}{2} = \frac{z - 2}{1}$ is $(α, β, γ)$ , then $14 (α + β + γ)$ is : [2024]

Q 16 :

The distance of the point (7, –2, 11) from the line $\frac{x - 6}{1} = \frac{y - 4}{0} = \frac{z - 8}{3}$ along the line $\frac{x - 5}{2} = \frac{y - 1}{- 3} = \frac{z - 5}{6}$ is : [2024]

Q 17 :

If the shortest distance between the lines $\frac{x - 4}{1} = \frac{y + 1}{2} = \frac{z}{- 3}$ and $\frac{x - λ}{2} = \frac{y + 1}{4} = \frac{z - 2}{- 5}$ is $\frac{6}{\sqrt{5}}$ , then the sum of all possible values of $λ$ is : [2024]

Q 20 :

Let P(3, 2, 3), Q(4, 6, 2) and R(7, 3, 2) be the vertices of $△ P Q R$ . Then the angle $∠ Q P R$ is [2024]

Q 21 :

Let $(α, β, γ)$ be the foot of the perpendicular from the point (1, 2, 3) on the line $\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3}$ . Then $19 (α + β + γ)$ is equal to [2024]

Q 24 :

Let $(α, β, γ)$ be the mirror image of the point (2, 3, 5) in the line $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ . Then $2 α + 3 β + 4 γ$ is equal to [2024]

Q 27 :

Consider a line L passing through the points P(1, 2, 1) and Q(2, 1, –1). If the mirror image of the point A(2, 2, 2) in the line L is $(α, β, γ)$ , then $α + β + 6 γ$ is equal to __________. [2024]

Q 28 :

Let the point $(- 1, α, β)$ lie on the line of the shortest distance between the lines $\frac{x + 2}{- 3} = \frac{y - 2}{4} = \frac{z - 5}{2}$ and $\frac{x + 2}{- 1} = \frac{y + 6}{2} = \frac{z - 1}{0}$ . Then ${(α - β)}^{2}$ is equal to _________. [2024]

Q 29 :

Let P be the point (10, –2, –1) and Q be the foot of the perpendicular drawn from the point R(1, 7, 6) on the line passing through the points (2, –5, 11) and (–6, 7, –5). Then the length of the line segment PQ is equal to __________. [2024]

Q 30 :

If the shortest distance between the lines $\frac{x - λ}{3} = \frac{y - 2}{- 1} = \frac{z - 1}{1}$ and $\frac{x + 2}{- 3} = \frac{y + 5}{2} = \frac{z - 4}{4}$ is $\frac{44}{\sqrt{30}}$ , then the largest possible value of $| λ |$ is equal to __________. [2024]