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

Mathematics · 3D Geometry

JEE Main 2026 — 5 April, Evening Shift — Question 38

Let a triangle $PQR$ be such that P and Q lie on the line $\frac{x+3}{8} = \frac{y-4}{2} = \frac{z+1}{2}$ and are at a distance of $6$ units from $R(1,2,3).$ If $(α,β,γ)$ is the centroid of $ΔPQR,$ then $α+β+γ$ is equal to:

Option A:
$4$
Option B:
$5$
Option C:
$6$
Correct
Option D:
$8$

Answer: C

Step-by-step solution

Let the point P is $(8 \lambda-3,2 \lambda+4,2 \lambda-1)$ $\mathrm{PR}=6 \Rightarrow \mathrm{PR}^{2}=36$ $(8 \lambda-4)^{2}+(2 \lambda+2)^{2}+(2 \lambda-4)^{2}=36$ $(4 \lambda-2)^{2}+(\lambda+1)^{2}+(\lambda-2)^{2}=9$ $\lambda^{2}-\lambda=0 \Rightarrow \lambda=0,1$ $\lambda=0 \Rightarrow \mathrm{P}(-3,4,-1)$ $\lambda=1 \Rightarrow P(5,6,1)$ $\mathrm{R}(1,2,3)$ $Q(1,4,1)$ $=\alpha+\beta+\gamma=6$

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

If the distance of the point (a,2,5) from the image of the point (1,2,7) in the line x-1/1 = y-2/1 = z-7/2 is 4, then the sum of all…

Answer: C

Step-by-step solution

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