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JEE Main 2026 · 5 Apr · Shift 2, Mathematics question 14

Let a triangle PQR be such that P and Q lie on the line x+38=y−42=z+12\dfrac{x+3}8=\dfrac{y-4}2=\dfrac{z+1}2 and are at a distance of 6 units from R(1,2,3)R(1,2,3). If (α,β,γ)(\alpha,\beta,\gamma) is the centroid of ΔPQR\Delta PQR, then α+β+γ\alpha+\beta+\gamma is equal to :

Choose one option
Answer and worked solution

Official answer: C

Foot of the perpendicular from R. On the line, A=(−3,4,−1)A=(-3,4,-1) and d⃗=(8,2,2)\vec d=(8,2,2). (R−A)⋅d⃗=(4,−2,4)⋅(8,2,2)=36(R-A)\cdot\vec d=(4,-2,4)\cdot(8,2,2)=36 and ∣d⃗∣2=72|\vec d|^2=72, so λ=12\lambda=\tfrac12 and the foot is M=(1,5,0)M=(1,5,0).

P and Q. They are equidistant from R, so they are symmetric about M, and P+Q=2MP+Q=2M.

Centroid.
2M+R3=(1,4,1),\frac{2M+R}3=(1,4,1),
so α+β+γ=6\alpha+\beta+\gamma=6.

(C)

Topics:Equation of a line in spaceAngles and distances
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