Mathematics · 3D Geometry

JEE Main 2026 — 5 April, Morning Shift — Question 36

The square of the distance of the point P(5,6,7) from the line (x-2)/2 = (y-5)/3 = (z-2)/4 is equal to:

Option A:
3
Option B:
5
Option C:
6
Correct
Option D:
8

Answer: C

Step-by-step solution

$\overrightarrow{\mathrm{PQ}} \cdot \overrightarrow{\mathrm{A}}=0$ $\Rightarrow(2 \lambda-3) 2+(3 \lambda-1) 3+(4 \lambda-5) 4=0$ $\Rightarrow \lambda=1$ Point $\mathrm{Q}=(4,8,6)$ $\mathrm{PQ}^{2}=6$

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Exam: JEE Main 2026
Paper: 5 April, Morning Shift
Subject: Mathematics
Chapter: 3D Geometry
Topic: Straight Lines in 3D Geometry

The square of the distance of the point of intersection of the lines r = (i+j-k) + λ(ai - j), a≠0 and r = (4i - k) + μ(2i +…

Answer: C

Step-by-step solution

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