Three Dimensional Geometry jee mains questions with solutions

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

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(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)

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(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]

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

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(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

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(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]

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

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(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]

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(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]

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[Figure]

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]

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(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]

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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]

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(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 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]

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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]

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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]

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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]

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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 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]

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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]

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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]

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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]

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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 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]