Let P and Q be the points on the line x+3/8=y-4/2=z+1/2 which are at a distance of 6 units from the point R (1,2,3) . If the centroid of the triangle PQR is (α, β, γ) , then α 2 +β 2 +γ 2 is:

Mathematics · 3D Geometry

JEE Main 2024 — 1 February, Shift 2 — Question 8

Let $P$ and $Q$ be the points on the line $\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}$ which are at a distance of 6 units from the point $\mathrm{R}(1,2,3)$ . If the centroid of the triangle PQR is $(\alpha, \beta, \gamma)$ , then $\alpha^{2}+\beta^{2}+\gamma^{2}$ is:

Option A:
26
Option B:
39
Option C:
18
Correct
Option D:
24

Answer: C

Step-by-step solution

$\mathrm{P}(8 \lambda-3,2 \lambda+4,2 \lambda-1)$

$\mathrm{PR}=6$

$(8 \lambda-4)^{2}+(2 \lambda+2)^{2}+(2 \lambda-4)^{2}=36$

$\lambda=0,1$

Hence $P(-3,4,-1) \& Q(5,6,1)$

Centroid of $\triangle \mathrm{PQR}=(1,4,1) \equiv(\alpha, \beta, \gamma)$

$\alpha^{2}+\beta^{2}+\gamma^{2}=18$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 1 February, Shift 2
Subject: Mathematics
Chapter: 3D Geometry
Topic: Vector & Cartesian forms of lines and planes

If the mirror image of the point P(3,4,9) in the line x-1/3=y+1/2=z-2/1 is (alpha,beta,gamma) , then 14(alpha+beta+gamma) is

Answer: C

Step-by-step solution

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