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 overrightarrow CD 2 is equal to :

Mathematics · 3D Geometry

JEE Main 2026 — 21 January, Evening Shift — Question 8

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 $\mathrm{L}_{3}$ is parallel to the vector $-3 \hat{i}+5 \hat{j}+16 \hat{k}$ and intersects the lines $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ at the points C and D , respectively, then $|\overrightarrow{\mathrm{CD}}|^{2}$ is equal to :

Option A:
$171$
Option B:
$290$
Correct
Option C:
$312$
Option D:
$89$

Answer: B

Step-by-step solution

$\quad \mathrm{L}_{1}: \frac{\mathrm{x}-2}{-3}=\frac{\mathrm{y}-6}{2}=\frac{\mathrm{z}-7}{4}$

Point C on $\mathrm{L}_{1}:\left(-3 \lambda_{1}+2,2 \lambda_{1}+6,4 \lambda_{1}+7\right)$

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

Point D on $\mathrm{L}_{2}:\left(2 \lambda_{2}+4, \lambda_{2}+3,3 \lambda_{2}+5\right)$

Dr's of line $L_{3}$ : $\mathrm{L}_{3}: \frac{2 \lambda_{2}+3 \lambda_{1}+2}{-3}=\frac{\lambda_{2}-2 \lambda_{1}-3}{5}=\frac{3 \lambda_{2}-4 \lambda_{1}-2}{16}$

$\lambda_{1}=-3, \lambda_{2}=2$

$C (11, 0, -5), D (8, 5, 11)$

$|\overrightarrow{\mathrm{CD}}|^{2}=3^{2}+5^{2}+16^{2}=290$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 21 January, Evening Shift
Subject: Mathematics
Chapter: 3D Geometry
Topic: Straight Lines in 3D Geometry

Let the line L pass through the point (-3,5,2) and make equal angles with the positive coordinate axes. If the distance of L from the point…

Answer: B

Step-by-step solution

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