Vectors & 3D GeometrymediumPYQ · JEE Main · 5 Apr 2026 · Shift 2 (Afternoon)Free

Vectors & 3D Geometry: Let Pqr Lie Distance Centroid Pqr Equals (JEE Main 2026)

JEE Maths question with a full step-by-step solution.

Question
Let △PQR\triangle PQR be such that PP and QQ lie on x+38=y−42=z+12\dfrac{x+3}{8}=\dfrac{y-4}{2}=\dfrac{z+1}{2} and are at distance 6 from R(1,2,3)R(1,2,3). If (α,β,γ)(\alpha,\beta,\gamma) is the centroid of △PQR\triangle PQR, then α+β+γ\alpha+\beta+\gamma equals:
A44
B55
C66correct
D88
Solution
Step 1: Parametrize points on the line A general point on the line is (8λ−3, 2λ+4, 2λ−1)(8\lambda-3,\ 2\lambda+4,\ 2\lambda-1). Step 2: Apply the distance condition from R(1,2,3)R(1,2,3)
(8λ−4)2+(2λ+2)2+(2λ−4)2=36(8\lambda-4)^2+(2\lambda+2)^2+(2\lambda-4)^2 = 36
64λ2−64λ+16+4λ2+8λ+4+4λ2−16λ+16=3664\lambda^2-64\lambda+16+4\lambda^2+8\lambda+4+4\lambda^2-16\lambda+16 = 36
72λ2−72λ=0  ⟹  λ(λ−1)=0  ⟹  λ=0 or λ=172\lambda^2-72\lambda = 0 \implies \lambda(\lambda-1)=0 \implies \lambda=0 \text{ or } \lambda=1
 Step 3: Determine PP and QQ
λ=0:  P=(−3,4,−1);λ=1:  Q=(5,6,1)\lambda=0:\; P=(-3,4,-1); \qquad \lambda=1:\; Q=(5,6,1)
 Step 4: Compute the centroid
G=(−3+5+13, 4+6+23, −1+1+33)=(1,4,1)G = \left(\frac{-3+5+1}{3},\ \frac{4+6+2}{3},\ \frac{-1+1+3}{3}\right) = (1,4,1)
α+β+γ=1+4+1=6\alpha+\beta+\gamma = 1+4+1 = 6
 Answer: (3)
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