Three Dimensional Geometry jee mains questions with solutions

Q 31 :

Let a line $l$ pass through the origin and be perpendicular to the lines, $l_{1} : \vec{r} = (\hat{i} - 11 \hat{j} - 7 \hat{k}) + λ (\hat{i} + 2 \hat{j} + 3 \hat{k}), λ \in ℝ$ and $l_{2} : \vec{r} = (- \hat{i} + \hat{k}) + μ (2 \hat{i} + 2 \hat{j} + \hat{k}), μ \in ℝ .$ If P is the point of intersection of $l$ and $l_{1}$ , and $Q (α, β, γ)$ is the foot of the perpendicular from $P$ on $l_{2}$ , then $9 (α + β + γ)$ is equal to _______ . [2023]

(5)

$Let l = (0 \hat{i} + 0 \hat{j} + 0 \hat{k}) + γ (a \hat{i} + b \hat{j} + c \hat{k}) = γ (a \hat{i} + b \hat{j} + c \hat{k})$

Since $l$ is perpendicular to the lines $l_{1}$ and $l_{2}$ ,

$a \hat{i} + b \hat{j} + c \hat{k} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 2 & 1 \end{matrix} |$ $= \hat{i} (2 - 6) - \hat{j} (1 - 6) + \hat{k} (2 - 4)$

$= - 4 \hat{i} + 5 \hat{j} - 2 \hat{k}$ $∴$ $l = γ (- 4 \hat{i} + 5 \hat{j} - 2 \hat{k})$

$P$ is the point of intersection of $l$ and $l_{1}$ .

$- 4 γ = 1 + λ, 5 γ = - 11 + 2 λ, - 2 γ = - 7 + 3 λ$

$By solving these equations, we get γ = - 1 .$

$So, P \equiv (4, - 5, 2)$

$Let Q (- 1 + 2 μ, 2 μ, 1 + μ)$

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

$\Rightarrow [(2 μ - 5) \hat{i} + (2 μ + 5) \hat{j} + (μ - 1) \hat{k}] \cdot (2 \hat{i} + 2 \hat{j} + \hat{k}) = 0$

$\Rightarrow 4 μ - 10 + 4 μ + 10 + μ - 1 = 0 \Rightarrow 9 μ = 1 \Rightarrow μ = \frac{1}{9}$

$Hence, 9 (α + β + γ) = 9 (\frac{- 7}{9} + \frac{2}{9} + \frac{10}{9}) = 5$

Q 32 :

Let the plane P contain the line $2 x + y - z - 3 = 0 = 5 x - 3 y + 4 z + 9$ and be parallel to the line $\frac{x + 2}{2} = \frac{3 - y}{- 4} = \frac{z - 7}{5} .$ Then the distance of the point $A (8, - 1, - 19)$ from the plane P measured parallel to the line $\frac{x}{- 3} = \frac{y - 5}{4} = \frac{2 - z}{- 12}$ is equal to _______ . [2023]

(26)

$Equation of the plane passing through the line of intersection of the given planes is$

$(2 x + y - z - 3) + λ (5 x - 3 y + 4 z + 9) = 0$

$\Rightarrow x (2 + 5 λ) + y (1 - 3 λ) + z (4 λ - 1) + 9 λ - 3 = 0 \dots (i)$

$The plane is parallel to the line$

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

$∴$ $2 (2 + 5 λ) + 4 (1 - 3 λ) + 5 (4 λ - 1) = 0$

$\Rightarrow 18 λ + 3 = 0 \Rightarrow λ = - \frac{1}{6}$

$Putting the value λ = - \frac{1}{6} in equation (i), we get$

$x (2 - \frac{5}{6}) + y (1 + \frac{3}{6}) + z (\frac{- 4}{6} - 1) - \frac{9}{6} - 3 = 0$

$\Rightarrow 7 x + 9 y - 10 z - 27 = 0$

$This is the required equation of plane.$

$Now, the given point is (8, - 1, - 19) and the equation of parallel line is \frac{x}{- 3} = \frac{y - 5}{4} = \frac{z - 2}{12}$

$∴$ $Equation of line passing through the given point and parallel to the line is$

$\frac{x - 8}{- 3} = \frac{y + 1}{4} = \frac{z + 19}{12} = k$

$\Rightarrow x = - 3 k + 8, y = 4 k - 1, z = 12 k - 19$

$This will satisfy the plane: 7 x + 9 y - 10 z - 27 = 0$

$- 21 k + 56 + 36 k - 9 - 120 k + 190 - 27 = 0$

$\Rightarrow - 105 k + 210 = 0 \Rightarrow k = 2$

$∴$ $x = 2, y = 7, z = 5$

$∴$ $Required distance = \sqrt{{(8 - 2)}^{2} + {(- 1 - 7)}^{2} + {(- 19 - 5)}^{2}}$

$= \sqrt{36 + 64 + 576} = \sqrt{676} = 26 units$

Q 33 :

Let a line L pass through the point P(2, 3, 1) and be parallel to the line $x + 3 y - 2 z - 2 = 0 = x - y + 2 z .$ If the distance of L from the point (5, 3, 8) is $α$ , then $3 α^{2}$ is equal to _________ . [2023]

(158)

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

$\vec{a} \times \vec{b} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & - 2 \\ 1 & - 1 & 2 \end{matrix} |$ $= 4 \hat{i} - 4 \hat{j} - 4 \hat{k} = 4 (\hat{i} - \hat{j} - \hat{k})$

$Line will be parallel to \vec{a} \times \vec{b} \Rightarrow \vec{n} = \hat{i} - \hat{j} - \hat{k}$

$Equation of the line passing through the point P (2, 3, 1) and parallel to \vec{a} \times \vec{b}$

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

${\vec{a}}_{1} = 2 \hat{i} + 3 \hat{j} + \hat{k}, {\vec{a}}_{2} = 5 \hat{i} + 3 \hat{j} + 8 \hat{k}$ $\Rightarrow {\vec{a}}_{2} - {\vec{a}}_{1} = 3 \hat{i} + 7 \hat{k}$

$({\vec{a}}_{2} - {\vec{a}}_{1}) \times \vec{n} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 0 & 7 \\ 1 & - 1 & - 1 \end{matrix} |$ $= 7 \hat{i} - \hat{j} (- 3 - 7) + \hat{k} (- 3)$

$= 7 \hat{i} + 10 \hat{j} - 3 \hat{k}$

$Now, α =$ $| \frac{({\vec{a}}_{2} - {\vec{a}}_{1}) \times \vec{n}}{| \vec{n} |} |$ $= \frac{| 7 \hat{i} + 10 \hat{j} - 3 \hat{k} |}{\sqrt{1 + 1 + 1}}$

$= \frac{\sqrt{49 + 100 + 9}}{\sqrt{3}} = \frac{\sqrt{158}}{\sqrt{3}} \Rightarrow 3 α^{2} = 158 units$

Q 34 :

The vertices B and C of a triangle ABC lie on the line $\frac{x}{1} = \frac{1 - y}{- 2} = \frac{z - 2}{3} .$ The coordinates of A and B are (1,6,3) and (4,9, $α$ ) respectively and C is at a distance of 10 units from B. The area (in sq. units) of $∆$ ABC is: [2026]

$5 \sqrt{13}$
$10 \sqrt{13}$
$20 \sqrt{13}$
$15 \sqrt{13}$

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(1)

$\frac{4}{1} = \frac{9 - 1}{2} = \frac{α - 2}{3} \Rightarrow α = 14$

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

$(λ - 1) \hat{i} + (2 λ - 5) \hat{j} + (3 λ - 1) \hat{k} = \vec{A D}$

$\Rightarrow λ - 1 + 4 λ - 10 + 9 λ - 3 = 0$

$\Rightarrow 14 λ = 14 \Rightarrow λ = 1$

$D = (1, 3, 5)$

$A D = \sqrt{3^{2} + 2^{2}} = \sqrt{13}$

$Ar (∆ A B C) = \frac{1}{2} \times \sqrt{13} \times 10 = 5 \sqrt{13}$

Q 35 :

Let the lines $L_{1} : \vec{r} = \hat{i} + 2 \hat{j} + 3 \hat{k} + λ (2 \hat{i} + 3 \hat{j} + 4 \hat{k}), λ \in ℝ$ and $L_{2} : \vec{r} = (4 \hat{i} + \hat{j}) + μ (5 \hat{i} + 2 \hat{j} + \hat{k}), μ \in ℝ,$ intersect at the point R. Let P and Q be the points lying on lines $L_{1}$ and $L_{2}$ respectively, such that $| \vec{P R} |$ $= \sqrt{29}$ and $| \vec{P Q} |$ $= \sqrt{\frac{47}{3}}$ . If the point P lies in the first octant, then $27 {(Q R)}^{2}$ is equal to [2026]

320
340
360
348

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(3)

$For POI$

$2 λ + 1 = 5 μ + 4; 3 λ + 2 = 2 μ + 1; 4 λ + 3 = μ$

$\Rightarrow λ = μ = - 1$

$R (- 1, - 1, - 1), P (2 λ + 1, 3 λ + 2, 4 λ + 3)$

$P R^{2} = 29 \Rightarrow {(2 λ + 2)}^{2} + {(3 λ + 3)}^{2} + {(4 λ + 4)}^{2} = 29$

$\Rightarrow λ = 0 or λ = - 2 (Reject)$

$\Rightarrow P (1, 2, 3)$

$Q (5 μ + 4, 2 μ + 1, μ)$

$| P Q |$ $= \sqrt{\frac{47}{3}} \Rightarrow P Q^{2} = \frac{47}{3}$

$\Rightarrow {(5 μ + 3)}^{2} + {(2 μ - 1)}^{2} + {(μ - 3)}^{2} = \frac{47}{3}$

$\Rightarrow μ = - \frac{1}{3}$

$Q = (\frac{7}{3}, \frac{1}{3}, - \frac{1}{3})$

${(Q R)}^{2} = {(\frac{7}{3} + 1)}^{2} + {(\frac{1}{3} + 1)}^{2} + {(- \frac{1}{3} + 1)}^{2}$

$= \frac{100 + 16 + 4}{9} = \frac{120}{9}$

$\Rightarrow 27 \times {(Q R)}^{2} = 27 \times \frac{120}{9} = 360$

Q 36 :

Let a line L passing through the point P(1,1,1) be perpendicular to the lines $\frac{x - 4}{4} = \frac{y - 1}{1} = \frac{z - 1}{1} and \frac{x - 17}{1} = \frac{y - 71}{1} = \frac{z}{0} .$ Let the line L intersect the $y z$ -plane at the point Q. Another line parallel to L and passing through the point S(1, 0, -1) intersects the $y z$ -plane at the point R. Then the square of the area of the parallelogram PQRS is equal to ________ . [2026]

(6)

$d_{1} = ⟨ 4, 1, 1 ⟩$ and $d_{2} = ⟨ 1, 1, 0 ⟩$

$d_{L} = d_{1} \times d_{2} =$ $| \begin{matrix} i & j & k \\ 4 & 1 & 1 \\ 1 & 1 & 0 \end{matrix} |$ $= ⟨ - 1, 1, 3 ⟩$

$Line L passes through P ⟨ 1, 1, 1 ⟩ with direction d_{L} = ⟨ - 1, 1, 3 ⟩$

$r (t) = ⟨ 1, 1, 1 ⟩ + t ⟨ - 1, 1, 3 ⟩ = ⟨ 1 - t, 1 + t, 1 + 3 t ⟩$

$For point Q, x = 0 \Rightarrow t = 1$

$Q = ⟨ 0, 2, 4 ⟩$

$Another line parallel to L passes through S ⟨ 1, 0, - 1 ⟩$

$r' (μ) = ⟨ 1, 0, - 1 ⟩ + μ ⟨ - 1, 1, 3 ⟩ = ⟨ 1 - μ, μ, - 1 + 3 μ ⟩$

$For point R, x = 0 \Rightarrow μ = 1$

$R = ⟨ 0, 1, 2 ⟩$

$Area of parallelogram with adjacent vectors \vec{P Q} and \vec{P S}$

$\vec{P Q} = ⟨ - 1, 1, 3 ⟩$

$\vec{P S} = ⟨ 0, - 1, - 2 ⟩$

Area of parallelogram

$\vec{P Q} \times \vec{P S} =$ $| \begin{matrix} i & j & k \\ - 1 & 1 & 3 \\ 0 & - 1 & - 2 \end{matrix} |$ $= ⟨ 1, - 2, 1 ⟩$

$Area = \sqrt{1^{2} + {(- 2)}^{2} + 1^{2}} = \sqrt{6}$

Q 37 :

If the distance of the point $P (43, α, β), β < 0,$ from the line $\vec{r} = 4 \hat{i} - \hat{k} + μ (2 \hat{i} + 3 \hat{k}), μ \in ℝ$ along a line with direction ratios $3, - 1, 0$ is $13 \sqrt{10},$ then $α^{2} + β^{2}$ is equal to _______ [2026]

(170)

$\frac{x - 43}{3} = \frac{y - α}{- 1} = \frac{z - β}{0} \Rightarrow P_{1} (43 + 3 λ, α - λ, β)$

$\frac{x - 4}{2} = \frac{y}{0} = \frac{z + 1}{3} \Rightarrow P_{1} (2 μ + 4, 0, 3 μ - 1)$

$∴$ $μ = \frac{3 λ + 39}{2}, α = λ, β = \frac{9 λ - 115}{2}$

$P (43, α, β), P_{1} (43 + 3 α, 0, β)$

${(P P_{1})}^{2} = 1690 = 10 α^{2} \Rightarrow α = 13, β = 1$

$∴$ $α^{2} + β^{2} = 170$

Q 38 :

The sum of all values of $α$ , for which the shortest distance between the lines

$\frac{x + 1}{α} = \frac{y - 2}{- 1} = \frac{z - 4}{- α} and \frac{x}{α} = \frac{y - 1}{2} = \frac{z - 1}{2 α}$ is $\sqrt{2}$ , is [2026]

8
- 6
6
- 8

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(2)

$\sqrt{2} =$ $\frac{| \begin{matrix} - 1 & 1 & 3 \\ α & - 1 & - α \\ α & 2 & 2 α \end{matrix} |}{| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ α & - 1 & - α \\ α & 2 & 2 α \end{matrix} |}$

$\sqrt{2} = \frac{- 1 (- 2 α + 2 α) - 1 (2 α^{2} + α^{2}) + 3 (2 α + α)}{| \hat{i} (- 2 α + 2 α) - \hat{j} (2 α^{2} + α^{2}) + \hat{k} (2 α + α) |}$

$\sqrt{2} = \frac{- 3 α^{2} + 9 α}{\sqrt{9 α^{4} + 9 α^{2}}}$

$\sqrt{2} = \frac{- α + 3}{\sqrt{α^{2} + 1}}$

$\Rightarrow 2 α^{2} + 2 = α^{2} + 9 - 6 α$

$α^{2} + 6 α - 7 = 0$

$(α + 7) (α - 1) = 0$

$α = - 7, 1$

$sum = - 7 + 1 = - 6$

option (2)

Q 39 :

If the image of the point $P (a, 2, a)$ in the line $\frac{x}{2} = \frac{y + a}{1} = \frac{z}{1}$ is Q and the image of Q in the line $\frac{x - 2 b}{2} = \frac{y - a}{1} = \frac{z + 2 b}{- 5}$ is P, then $a + b$ is equal to ______. [2026]

(3)

$\frac{2 λ - a + a}{2} = - 5 μ - 2 b$

$and \frac{a + 4 λ - a}{2} = 2 μ + 2 b$

$and \frac{2 λ - 2 a - 2 + 2}{2} = μ + a$

$\Rightarrow λ - μ = b$

$λ - μ = 2 a$

$λ + 5 μ = - 2 b$

$\Rightarrow b = 2 a and λ = a, μ = - a$

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

$2 a - 2 = 0$

$a = 1$

$b = 2 a = 2 \times 1 = 2$

$a + b = 1 + 2 = 3$

Q 40 :

Let the line $L_{1}$ be parallel to the vector $- 3 \hat{i} + 2 \hat{j} + 4 \hat{k}$ and pass through the point (2,6,7), and the line $L_{2}$ be parallel to the vector $2 \hat{i} + \hat{j} + 3 \hat{k}$ and pass through the point (4,3,5). If the line $L_{3}$ is parallel to the vector $- 3 \hat{i} + 5 \hat{j} + 16 \hat{k}$ and intersects the lines $L_{1} and L_{2}$ at the points C and D, respectively, then $| \vec{C D} |^{2}$ is equal to: [2026]

171
89
290
312

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(3)

$L_{1} : \frac{x - 2}{- 3} = \frac{y - 6}{2} = \frac{z - 7}{4}$

Point $C$ on $L_{1}$ : $(- 3 λ_{1} + 2, 2 λ_{1} + 6, 4 λ_{1} + 7)$

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

Point $D$ on $L_{2}$ : $(2 λ_{2} + 4, λ_{2} + 3, 3 λ_{2} + 5)$

D.R's of line $L_{3}$ :

$L_{3} : \frac{2 λ_{2} + 3 λ_{1} + 2}{- 3} = \frac{λ_{2} - 2 λ_{1} - 3}{5} = \frac{3 λ_{2} - 4 λ_{1} - 2}{16}$

$λ_{1} = - 3, λ_{2} = 2$

$C (11, 0, - 5)$

$D (8, 5, 11)$

$| \vec{C D} |^{2} = 3^{2} + 5^{2} + 16^{2} = 290$

Q 41 :

Let L be the line $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z + 3}{6}$ and let S be the set of all points (a,b,c) on L, whose distance from the line $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z - 9}{0}$ along the line L is 7. Then $\sum_{(a, b, c) \in S} (a + b + c)$ is equal to: [2026]

6
40
34
28

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(3)

$M is the point of intersection of L_{1} and L_{2}$

$\Rightarrow 2 λ - 1 = 2 μ - 1, 3 λ - 1 = 3 μ - 1, 6 λ - 3 = 9$

$\Rightarrow λ = 2 = μ$

$\Rightarrow M (3, 5, 9)$

$Now let point P be (2 K - 1, 3 K - 1, 6 K - 3) on L_{2},$ $such that P M = 7$

$\Rightarrow \sqrt{{(2 K - 4)}^{2} + {(3 K - 6)}^{2} + {(6 K - 12)}^{2}} = 7$

$\Rightarrow 49 K^{2} + 196 - 196 K = 49$

$\Rightarrow K^{2} + 4 - 4 K = 1$

$\Rightarrow K^{2} - 4 K + 3 = 0$

$\Rightarrow K = 1, 3$

$So points P and Q are (1, 2, 3) and (5, 8, 15)$

$So sum of all coordinates of P and Q = 34$

Q 42 :

If the distances of the point (1,2,a) from the line $\frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{1}$ along the lines $L_{1} : \frac{x - 1}{3} = \frac{y - 2}{4} = \frac{z - a}{b} and L_{2} : \frac{x - 1}{1} = \frac{y - 2}{4} = \frac{z - a}{c}$ are equal, then $a + b + c$ is equal to [2026]

5
6
7
4

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(3)

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

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

$L_{2} : \frac{x - 1}{1} = \frac{y - 2}{4} = \frac{z - a}{c} = μ$

$Let A (3 λ + 1, 4 λ + 2, b λ + a)$

$It lies on L$

$∴$ $\frac{3 λ}{1} = \frac{4 λ + 2}{2} = \frac{b λ + a - 1}{1}$

$\Rightarrow λ = 1 and a + b - 1 = 3$

$\Rightarrow A (4, 6, 4), a + b = 4 . . . (1)$

$Let B (μ + 1, 4 μ + 2, c μ + a)$

$It also lies on L$

$\frac{μ}{1} = \frac{4 μ + 2}{2} = \frac{c μ + a - 1}{1}$

$\Rightarrow 2 μ = 4 μ + 2$

$\Rightarrow μ = - 1$

$a - c - 1 = - 1$

$\Rightarrow a = c . . . (2), B (0, - 2, 0)$

$Also P A = P B, P (1, 2, a), A (4, 6, 4)$

$\Rightarrow 9 + 16 + {(a - 4)}^{2} = 1 + 16 + a^{2}$

$\Rightarrow 16 + 8 = 8 a$

$\Rightarrow a = 3$ $∴$ $c = 3, b = 1$

$∴$ $a + b + c = 7$

Q 43 :

Let $(α, β, γ)$ be the mirror image of $(1, - 2, 3)$ in the line $\frac{x + 1}{2} = \frac{y - 3}{4} = \frac{z + 2}{3}$ then $2 α + 4 β + 3 γ + 2 =____$

5
11
9
10

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(1)

$A B ⊥ line, (α - 1) 2 + 4 (β + 2) + 3 (γ - 3) = 0$

$∴$ $2 α + 4 β + 3 γ = 3$

Q 44 :

List-I List-II

(A) If the line $\frac{x - 4}{1} = \frac{y - 2}{1} = \frac{z - 2 k}{2}$ lies in the plane
$2 x - 4 y + z = 3$ , then $k$ is equal to (I) $\frac{9}{2}$

(B) If the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and
$\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect, then value of $k$ is (II) $\frac{3}{2}$

(C) If a plane passes through the point (1,1,1) cuts the coordinate axes at A, B & C respectively such that OA = OB = OC where O is the origin, then volume of tetrahedron OABC is equal to (III) $\frac{5}{2}$

(D) If a plane P passes through (1, -2, 1) is perpendicular to two planes
$2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ . The distance of plane $P$ from the point $(- 1, \frac{5}{\sqrt{2}}, \frac{3}{\sqrt{2}})$ is (IV) $\frac{7}{2}$

(A) - II, (B) - IV, (C) - I, (D) - III
(A) - I, (B) - III, (C) - IV, (D) - II
(A) - IV, (B) - II, (C) - I, (D) - III
(A) - II, (B) - I, (C) - III, (D) - IV

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(1)

(A) As line is parallel to the plane, the point $(4, 2, 2 k)$ should lie in the given plane

$∴$ $2 (4) - 4 (2) + 2 k = 3 \Rightarrow k = \frac{3}{2}$

(B) Any point on given lines are $(2 λ + 1, 3 λ - 1, 4 λ + 1)$ and $(μ + 3, 2 μ + k, μ)$ respectively.

By equating the corresponding coordinates $\Rightarrow k = \frac{9}{2}$

(C) Equation of plane is $\frac{x}{a} + \frac{y}{a} + \frac{z}{a} = 1$ satisfy it by given point $\Rightarrow \frac{1}{a} + \frac{1}{a} + \frac{1}{a} = 1$ $\Rightarrow a = 3$

Hence A(3,0,0), B(0,3,0), C(0,0,3) $\Rightarrow$ volume of tetrahedron $= \frac{1}{6}$ $| \begin{matrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{matrix} |$ $= \frac{9}{2} cubic units$

(D) The equation of plane is $a (x - 1) + b (y + 2) + c (z - 1) = 0$ and $2 a - 2 b + c = 0$

$a - b + 2 c = 0$ $\Rightarrow \frac{a}{1} = \frac{b}{1} = \frac{c}{0}$ Hence equation of plane is $x + y + 1 = 0$

Q 45 :

The vector equation of the plane through the line of intersection of the planes $x + y + z = 1$ and $2 x + 3 y + 4 z = 5$ which is perpendicular to the plane $x - y + z = 0$ is:

$\vec{r} \times (\hat{i} + \hat{k}) + 2 = 0$
$\vec{r} \cdot (\hat{i} - \hat{k}) - 2 = 0$
$\vec{r} \cdot (\hat{i} - \hat{k}) + 2 = 0$
$\vec{r} \times (\hat{i} - \hat{k}) + 2 = 0$

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(3)

$Let the plane be$ $(x + y + z - 1) + λ (2 x + 3 y + 4 z - 5) = 0$

$\Rightarrow (2 λ + 1) x + (3 λ + 1) y + (4 λ + 1) z - (5 λ + 1) = 0$

$⊥ to the plane$ $x - y + z = 0$

$\Rightarrow λ = - \frac{1}{3}$

$\Rightarrow the required plane is x - z + 2 = 0$

Q 46 :

Which of the following must be the true value of the statements below in that order.

Statement I: For square matrices A and B of same order (with real entries) (O is null matrix and $I$ is Identity matrix of same order as A and B). If AB = O then ${(\det A)}^{2} + {(\det B)}^{2} = 0$

Statement II: If $l_{i}, m_{i}, n_{i}$ $(i = 1, 2, 3)$ denote the direction cosines of 3 mutual perpendicular vectors then $A = (\begin{matrix} l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2} \\ l_{3} & m_{3} & n_{3} \end{matrix})$ is an orthogonal matrix.

Statement III: For two distinct lines with shortest distance $d \neq 0$ we can have a point A on line one and point B on line two such that AB = d then A and B be unique points on them in all possible cases.

FFT
FFF
FTF
FTT

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(3)

$A B = 0,$ $| B |$ $\neq 0,$ $A B = 0 \Rightarrow A = 0$

Q 47 :

Let Q be the cube with the set of vertices ${(x_{1}, x_{2}, x_{3}) \in R^{3} : x_{1}, x_{2}, x_{3} \in {0, 1}}$ . Let F be the set of all twelve lines containing the diagonals of the six faces of the cube Q. Let S be the set of all four lines containing the main diagonals of the cube Q; for instance, the line passing through the vertices (0,0,0) and (1,1,1) is in S. For lines $l_{1}$ and $l_{2}$ , let $d (l_{1}, l_{2})$ denote the shortest distance between them. Then the maximum value of $d (l_{1}, l_{2})$ , as $l_{1}$ varies over F and $l_{2}$ varies over S, is $λ$ . Find the value of $λ^{- 2}$ .

(6)

$Equation of face diagonal OD line is$ $l_{1} : \vec{r} = λ (\hat{i} + \hat{j})$

$Equation of main diagonal BE is$ $l_{2} : \vec{r} = \hat{j} + μ (\hat{i} - \hat{j} + \hat{k})$

$Shortest distance =$ $| \frac{\hat{j} \cdot (\hat{i} + \hat{j}) \times (\hat{i} - \hat{j} + \hat{k})}{(\hat{i} + \hat{j}) \times (\hat{i} - \hat{j} + \hat{k})} |$

$=$ $| \frac{\hat{j} \cdot (\hat{i} - \hat{j} - 2 \hat{k})}{\hat{i} - \hat{j} - 2 \hat{k}} |$ $= \frac{1}{\sqrt{6}}$

$In other case S.D is zero.$

Q 48 :

The length of perpendicular drawn from P(1, 2, 3) to the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}$ is

9
14
7
21

1

2

3

4

(3)

$Let N be (3 λ + 6, 2 λ + 7, 2 λ + 7) such that P N is perpendicular to the line$

$Then λ = - 1$

$∴$ $N = (3, 5, 9)$

$∴$ $P N = 7$

Q 49 :

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

16
6
12
15

1

2

3

4

(1)

${\vec{a}}_{1} = (1, 2, 3),$ ${\vec{a}}_{2} = (2, 4, 5)$

${\vec{b}}_{1} = (2, 3, λ),$ ${\vec{b}}_{2} = (1, 4, 5)$

$Shortest distance:$

$\frac{| ({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2}) |}{| {\vec{b}}_{1} \times {\vec{b}}_{2} |}$ $= \frac{1}{\sqrt{3}}$

${\vec{a}}_{2} - {\vec{a}}_{1} = (1, 2, 2)$

${\vec{b}}_{1} \times {\vec{b}}_{2} = (15 - 4 λ, λ - 10, 5)$

$So,$

$\frac{| 5 - 2 λ |}{\sqrt{17 λ^{2} - 140 λ + 350}} = \frac{1}{\sqrt{3}}$

$Squaring:$

$3 {(5 - 2 λ)}^{2} = 17 λ^{2} - 140 λ + 350$

$λ^{2} - 16 λ + 55 = 0$

$(λ - 5) (λ - 11) = 0$

$λ = 5, 11$

$Sum = 16$

Q 50 :

If the length of the perpendicular drawn from the point $(a, 4, 2), a > 0$ on the line $\frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z - 1}{- 1}$ is $2 \sqrt{6}$ units and $Q (α_{1}, α_{2}, α_{3})$ is the image of the point P on this line, then $a + \sum_{i = 1}^{3} α_{i}$ is equal to

(8)

$Given line \frac{X + 1}{2} = \frac{Y - 3}{3} = \frac{Z - 1}{- 1} = λ$

$X = 2 λ - 1, Y = 3 λ + 3, Z = - λ + 1$

$(2 λ - 1 - a) 2 + (3 λ - 1) 3 + (- λ - 1) (- 1) = 0$

$\Rightarrow 4 λ - 2 - 2 a + 9 λ - 3 + λ + 1 = 0$

$14 λ - 4 - 2 a = 0$ $\Rightarrow 7 λ - 2 - a = 0$

$\Rightarrow {(5 λ - 1)}^{2} + {(3 λ - 1)}^{2} + {(λ - 1)}^{2} = 24$

$35 λ^{2} - 14 λ - 21 = 0$ $\Rightarrow (λ - 1) (35 λ + 21) = 0$

$For, λ = 1 \Rightarrow a = 5$

$Let (α_{1}, α_{2}, α_{3}) be reflection point P$

$α_{1} + 5, α_{2} + 4 = 12, α_{3} + 2 = 0$

$α_{1} = - 3, α_{2} = 8, α_{3} = - 2$

$α_{1} + α_{2} + α_{3} = 8$

Q 51 :

A straight line L intersects perpendicularly both the lines:

$\frac{x + 2}{2} = \frac{y + 6}{3} = \frac{z - 34}{- 10}$ and $\frac{x + 6}{4} = \frac{y - 7}{- 3} = \frac{z - 7}{- 2}$

Then the square of perpendicular distance of origin from L is

5
6
7
8

1

2

3

4

(1)

Line L is the shortest distance line of given lines.

Q 52 :

Let A be a point on the line $\vec{r} = (1 - 3 μ) \hat{i} + (μ - 1) \hat{j} + (2 + 5 μ) \hat{k}$ and B = (3,2,6) be a point in the space. Then the value of $' 4 μ'$ for which the vector $\bar{A B}$ is parallel to plane $x - 4 y + 3 z = 1$ is:

$1$
$\frac{1}{8}$
$\frac{1}{2}$
$- 1$

1

2

3

4

(1)

$Let A is (1 - 3 μ, μ - 1, 2 + 5 μ)$

$\bar{A B} = (3 μ + 2) \hat{i} + (3 - μ) \hat{j} + (4 - 5 μ) \hat{k} which is parallel to plane x - 4 y + 3 z = 1$

$- 8 μ + 2 = 0 \Rightarrow μ = \frac{1}{4}$ $∴$ $4 μ = 1$

Q 53 :

Twice the distance of the point (1, 0, 2) from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 16$ is

$2 \sqrt{14}$
$8$
$3 \sqrt{21}$
$26$

1

2

3

4

(4)

$\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12} = λ [say]$ $. . . (i)$

$And equation of plane is x - y + z = 16$

$Any point on the line (i) is (3 λ + 2, 4 λ - 1, 12 λ + 2)$

$Let this point of intersection of the line and plane.$

$(3 λ + 2) - (4 λ - 1) + (12 λ + 2) = 16$

$∴$ $11 λ = 11$ $\Rightarrow λ = 1$

$So, the point of intersection is (5, 3, 14)$

$Now, distance between the points (1, 0, 2) and (5, 3, 14)$

$= \sqrt{{(5 - 1)}^{2} + {(3 - 0)}^{2} + {(14 - 2)}^{2}}$

$= \sqrt{16 + 9 + 144}$

$= \sqrt{169}$

$= 13$

$∴$ $Twice the distance = 26$

Q 54 :

If the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{α}$ and $\frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z}{β}$ intersect, then the absolute value of $(β - α)$ is _______

3
- 3
13
- 13

1

2

3

4

(1)

Given the lines $\frac{x - 1}{2} = \frac{2 - y}{- 3} = \frac{z - 3}{α}$ and $\frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z}{β}$ intersect

So, point on first line (1,2,3) and point on second line (4,1,0)

Vector joining both points is $- 3 \hat{i} + \hat{j} + 3 \hat{k}$

Now vector along first line is $2 \hat{i} + 3 \hat{j} + α \hat{k}$

Also vector along second line is $5 \hat{i} + 2 \hat{j} + β \hat{k}$

Now these three vectors must be coplanar $\Rightarrow$ $| \begin{matrix} 2 & 3 & α \\ 5 & 2 & β \\ - 3 & 1 & 3 \end{matrix} |$ $= 0$

$\Rightarrow 2 (6 - β) - 3 (15 + 3 β) + α (11) = 0$

$\Rightarrow α - β = 3$

Q 55 :

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 sum of all possible values of $λ$ is _____

(8)

$| \frac{(\vec{a} - \vec{c}) \cdot (\vec{b} - \vec{d})}{| \vec{b} \times \vec{d} |} |$ $= \frac{6}{\sqrt{5}}$

Topic Question Set

Q 33 :

Let a line L pass through the point P(2, 3, 1) and be parallel to the line $x + 3 y - 2 z - 2 = 0 = x - y + 2 z .$ If the distance of L from the point (5, 3, 8) is $α$ , then $3 α^{2}$ is equal to _________ . [2023]

Q 34 :

The vertices B and C of a triangle ABC lie on the line $\frac{x}{1} = \frac{1 - y}{- 2} = \frac{z - 2}{3} .$ The coordinates of A and B are (1,6,3) and (4,9, $α$ ) respectively and C is at a distance of 10 units from B. The area (in sq. units) of $∆$ ABC is: [2026]

Q 37 :

If the distance of the point $P (43, α, β), β < 0,$ from the line $\vec{r} = 4 \hat{i} - \hat{k} + μ (2 \hat{i} + 3 \hat{k}), μ \in ℝ$ along a line with direction ratios $3, - 1, 0$ is $13 \sqrt{10},$ then $α^{2} + β^{2}$ is equal to _______ [2026]

Q 38 :

The sum of all values of $α$ , for which the shortest distance between the lines

$\frac{x + 1}{α} = \frac{y - 2}{- 1} = \frac{z - 4}{- α} and \frac{x}{α} = \frac{y - 1}{2} = \frac{z - 1}{2 α}$ is $\sqrt{2}$ , is [2026]

Q 39 :

If the image of the point $P (a, 2, a)$ in the line $\frac{x}{2} = \frac{y + a}{1} = \frac{z}{1}$ is Q and the image of Q in the line $\frac{x - 2 b}{2} = \frac{y - a}{1} = \frac{z + 2 b}{- 5}$ is P, then $a + b$ is equal to ______. [2026]

Q 41 :

Let L be the line $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z + 3}{6}$ and let S be the set of all points (a,b,c) on L, whose distance from the line $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z - 9}{0}$ along the line L is 7. Then $\sum_{(a, b, c) \in S} (a + b + c)$ is equal to: [2026]

Q 42 :

If the distances of the point (1,2,a) from the line $\frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{1}$ along the lines $L_{1} : \frac{x - 1}{3} = \frac{y - 2}{4} = \frac{z - a}{b} and L_{2} : \frac{x - 1}{1} = \frac{y - 2}{4} = \frac{z - a}{c}$ are equal, then $a + b + c$ is equal to [2026]

Q 43 :

Let $(α, β, γ)$ be the mirror image of $(1, - 2, 3)$ in the line $\frac{x + 1}{2} = \frac{y - 3}{4} = \frac{z + 2}{3}$ then $2 α + 4 β + 3 γ + 2 =____$

Q 45 :

The vector equation of the plane through the line of intersection of the planes $x + y + z = 1$ and $2 x + 3 y + 4 z = 5$ which is perpendicular to the plane $x - y + z = 0$ is:

Q 48 :

The length of perpendicular drawn from P(1, 2, 3) to the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}$ is

Q 49 :

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

Q 50 :

If the length of the perpendicular drawn from the point $(a, 4, 2), a > 0$ on the line $\frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z - 1}{- 1}$ is $2 \sqrt{6}$ units and $Q (α_{1}, α_{2}, α_{3})$ is the image of the point P on this line, then $a + \sum_{i = 1}^{3} α_{i}$ is equal to

Q 51 :

A straight line L intersects perpendicularly both the lines:

$\frac{x + 2}{2} = \frac{y + 6}{3} = \frac{z - 34}{- 10}$ and $\frac{x + 6}{4} = \frac{y - 7}{- 3} = \frac{z - 7}{- 2}$

Then the square of perpendicular distance of origin from L is

Q 52 :

Let A be a point on the line $\vec{r} = (1 - 3 μ) \hat{i} + (μ - 1) \hat{j} + (2 + 5 μ) \hat{k}$ and B = (3,2,6) be a point in the space. Then the value of $' 4 μ'$ for which the vector $\bar{A B}$ is parallel to plane $x - 4 y + 3 z = 1$ is:

Q 53 :

Twice the distance of the point (1, 0, 2) from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 16$ is

Q 54 :

If the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{α}$ and $\frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z}{β}$ intersect, then the absolute value of $(β - α)$ is _______

Q 55 :

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 sum of all possible values of $λ$ is _____

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	List-I		List-II
(A)	If the line $\frac{x - 4}{1} = \frac{y - 2}{1} = \frac{z - 2 k}{2}$ lies in the plane $2 x - 4 y + z = 3$ , then $k$ is equal to	(I)	$\frac{9}{2}$
(B)	If the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect, then value of $k$ is	(II)	$\frac{3}{2}$
(C)	If a plane passes through the point (1,1,1) cuts the coordinate axes at A, B & C respectively such that OA = OB = OC where O is the origin, then volume of tetrahedron OABC is equal to	(III)	$\frac{5}{2}$
(D)	If a plane P passes through (1, -2, 1) is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ . The distance of plane $P$ from the point $(- 1, \frac{5}{\sqrt{2}}, \frac{3}{\sqrt{2}})$ is	(IV)	$\frac{7}{2}$

Topic Question Set

Q 32 : Let the plane P contain the line 2x+y-z-3=0=5x-3y+4z+9 and be parallel to the line x+22=3-y-4=z-75. Then the distance of the point A(8,-1,-19) from the plane P measured parallel to the line x-3=y-54=2-z-12 is equal to _______ . [2023]

Q 33 : Let a line L pass through the point P(2, 3, 1) and be parallel to the line x+3y-2z-2=0=x-y+2z. If the distance of L from the point (5, 3, 8) is α, then 3α2 is equal to _________ . [2023]

Q 34 : The vertices B and C of a triangle ABC lie on the line x1=1-y-2=z-23. The coordinates of A and B are (1,6,3) and (4,9,α) respectively and C is at a distance of 10 units from B. The area (in sq. units) of ∆ABC is: [2026]

Q 37 : If the distance of the point P(43,α,β), β<0, from the line r→=4i^-k^+μ(2i^+3k^), μ∈ℝ along a line with direction ratios 3, -1, 0 is 1310, then α2+β2 is equal to _______ [2026]

Q 38 : The sum of all values of α, for which the shortest distance between the lines x+1α=y-2-1=z-4-α and xα=y-12=z-12α is 2, is [2026]

Q 39 : If the image of the point P(a,2,a) in the line x2=y+a1=z1 is Q and the image of Q in the line x-2b2=y-a1=z+2b-5 is P, then a+b is equal to ______. [2026]

Q 41 : Let L be the line x+12=y+13=z+36 and let S be the set of all points (a,b,c) on L, whose distance from the line x+12=y+13=z-90 along the line L is 7. Then ∑(a,b,c)∈S(a+b+c) is equal to: [2026]

Q 42 : If the distances of the point (1,2,a) from the line x-11=y2=z-11 along the lines L1: x-13=y-24=z-ab and L2: x-11=y-24=z-ac are equal, then a+b+c is equal to [2026]

Q 43 : Let (α,β,γ) be the mirror image of (1,-2,3) in the line x+12=y-34=z+23 then 2α+4β+3γ+2=____

Q 45 : The vector equation of the plane through the line of intersection of the planes x+y+z=1 and 2x+3y+4z=5 which is perpendicular to the plane x-y+z=0 is:

Q 48 : The length of perpendicular drawn from P(1, 2, 3) to the line x-63=y-72=z-7-2 is

Q 49 : If the shortest distance between the lines x-12=y-23=z-3λ and x-21=y-44=z-55 is 13, then the sum of all possible values of λ is :

Q 50 : If the length of the perpendicular drawn from the point (a,4,2), a>0 on the line x+12=y-33=z-1-1 is 26 units and Q(α1,α2,α3) is the image of the point P on this line, then a+∑i=13αi is equal to

Q 51 : A straight line L intersects perpendicularly both the lines: x+22=y+63=z-34-10 and x+64=y-7-3=z-7-2 Then the square of perpendicular distance of origin from L is

Q 52 : Let A be a point on the line r→=(1-3μ)i^+(μ-1)j^+(2+5μ)k^ and B = (3,2,6) be a point in the space. Then the value of '4μ' for which the vector AB¯ is parallel to plane x-4y+3z=1 is:

Q 53 : Twice the distance of the point (1, 0, 2) from the point of intersection of the line x-23=y+14=z-212 and the plane x-y+z=16 is

Q 54 : If the lines x-12=y-23=z-3α and x-45=y-12=zβ intersect, then the absolute value of (β-α) is _______

Q 55 : If the shortest distance between the lines x-41=y+12=z-3 and x-λ2=y+14=z-2-5 is 65. Then sum of all possible values of λ is _____

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Q 33 :

Let a line L pass through the point P(2, 3, 1) and be parallel to the line $x + 3 y - 2 z - 2 = 0 = x - y + 2 z .$ If the distance of L from the point (5, 3, 8) is $α$ , then $3 α^{2}$ is equal to _________ . [2023]

Q 34 :

The vertices B and C of a triangle ABC lie on the line $\frac{x}{1} = \frac{1 - y}{- 2} = \frac{z - 2}{3} .$ The coordinates of A and B are (1,6,3) and (4,9, $α$ ) respectively and C is at a distance of 10 units from B. The area (in sq. units) of $∆$ ABC is: [2026]

Q 37 :

If the distance of the point $P (43, α, β), β < 0,$ from the line $\vec{r} = 4 \hat{i} - \hat{k} + μ (2 \hat{i} + 3 \hat{k}), μ \in ℝ$ along a line with direction ratios $3, - 1, 0$ is $13 \sqrt{10},$ then $α^{2} + β^{2}$ is equal to _______ [2026]

Q 38 :

The sum of all values of $α$ , for which the shortest distance between the lines

$\frac{x + 1}{α} = \frac{y - 2}{- 1} = \frac{z - 4}{- α} and \frac{x}{α} = \frac{y - 1}{2} = \frac{z - 1}{2 α}$ is $\sqrt{2}$ , is [2026]

Q 39 :

If the image of the point $P (a, 2, a)$ in the line $\frac{x}{2} = \frac{y + a}{1} = \frac{z}{1}$ is Q and the image of Q in the line $\frac{x - 2 b}{2} = \frac{y - a}{1} = \frac{z + 2 b}{- 5}$ is P, then $a + b$ is equal to ______. [2026]

Q 41 :

Let L be the line $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z + 3}{6}$ and let S be the set of all points (a,b,c) on L, whose distance from the line $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z - 9}{0}$ along the line L is 7. Then $\sum_{(a, b, c) \in S} (a + b + c)$ is equal to: [2026]

Q 42 :

If the distances of the point (1,2,a) from the line $\frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{1}$ along the lines $L_{1} : \frac{x - 1}{3} = \frac{y - 2}{4} = \frac{z - a}{b} and L_{2} : \frac{x - 1}{1} = \frac{y - 2}{4} = \frac{z - a}{c}$ are equal, then $a + b + c$ is equal to [2026]

Q 43 :

Let $(α, β, γ)$ be the mirror image of $(1, - 2, 3)$ in the line $\frac{x + 1}{2} = \frac{y - 3}{4} = \frac{z + 2}{3}$ then $2 α + 4 β + 3 γ + 2 =____$

Q 45 :

The vector equation of the plane through the line of intersection of the planes $x + y + z = 1$ and $2 x + 3 y + 4 z = 5$ which is perpendicular to the plane $x - y + z = 0$ is:

Q 48 :

The length of perpendicular drawn from P(1, 2, 3) to the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}$ is

Q 49 :

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

Q 50 :

If the length of the perpendicular drawn from the point $(a, 4, 2), a > 0$ on the line $\frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z - 1}{- 1}$ is $2 \sqrt{6}$ units and $Q (α_{1}, α_{2}, α_{3})$ is the image of the point P on this line, then $a + \sum_{i = 1}^{3} α_{i}$ is equal to

Q 51 :

A straight line L intersects perpendicularly both the lines:

$\frac{x + 2}{2} = \frac{y + 6}{3} = \frac{z - 34}{- 10}$ and $\frac{x + 6}{4} = \frac{y - 7}{- 3} = \frac{z - 7}{- 2}$

Then the square of perpendicular distance of origin from L is

Q 52 :

Let A be a point on the line $\vec{r} = (1 - 3 μ) \hat{i} + (μ - 1) \hat{j} + (2 + 5 μ) \hat{k}$ and B = (3,2,6) be a point in the space. Then the value of $' 4 μ'$ for which the vector $\bar{A B}$ is parallel to plane $x - 4 y + 3 z = 1$ is:

Q 53 :

Twice the distance of the point (1, 0, 2) from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 16$ is

Q 54 :

If the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{α}$ and $\frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z}{β}$ intersect, then the absolute value of $(β - α)$ is _______

Q 55 :

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 sum of all possible values of $λ$ is _____