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JEE Main 202621 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2026 · 21 January · Evening Shift · Mathematics · Three Dimensional Geometry

Let the line L_ 1 be parallel to the vector -3 i +2 j +4 k and pass...

Let the line L1\mathrm{L}_{1} be parallel to the vector −3i^+2j^+4k^-3 \hat{i}+2 \hat{j}+4 \hat{k} and pass through the point (2,6,7), and the line L2\mathrm{L}_{2} be parallel to the vector 2i^+j^+3k^2 \hat{i}+\hat{j}+3 \hat{k} and pass through the point (4,3,5)(4,3,5). If the line L3\mathrm{L}_{3} is parallel to the vector −3i^+5j^+16k^-3 \hat{i}+5 \hat{j}+16 \hat{k} and intersects the lines L1\mathrm{L}_{1} and L2\mathrm{L}_{2} at the points C and D, respectively, then ∣CD→∣2|\overrightarrow{C D}|^{2} is equal to :

Options

  1. A89
  2. B312
  3. C171
  4. D290

Correct answer

D. 290

Step-by-step solution

L1:x−2−3=y−62=z−74L_1 : \dfrac{x-2}{-3} = \dfrac{y-6}{2} = \dfrac{z-7}{4}
Point C on L1L_1: (−3λ1+2,2λ1+6,4λ1+7)(-3\lambda_1 + 2, 2\lambda_1 + 6, 4\lambda_1 + 7)
L2:x−42=y−31=z−53L_2 : \dfrac{x-4}{2} = \dfrac{y-3}{1} = \dfrac{z-5}{3}
Point D on L2L_2: (2λ2+4,λ2+3,3λ2+5)(2\lambda_2 + 4, \lambda_2 + 3, 3\lambda_2 + 5)
Dr's of line L3L_3 :
L3:2λ2+3λ1+2−3=λ2−2λ1−35=3λ2−4λ1−216L_3 : \dfrac{2\lambda_2 + 3\lambda_1 + 2}{-3} = \dfrac{\lambda_2 - 2\lambda_1 - 3}{5} = \dfrac{3\lambda_2 - 4\lambda_1 - 2}{16}
λ1=−3,λ2=2\lambda_1 = -3, \lambda_2 = 2
C(11,0,−5)C(11, 0, -5)
D(8,5,11)D(8, 5, 11)
∣CD→∣2=32+52+162=290|\overrightarrow{CD}|^2 = 3^2 + 5^2 + 16^2 = 290
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