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 $\mathrm{R}(1,2,3)$. If $(\alpha, \beta, \gamma)$ is the centroid of $\Delta \mathrm{PQR}$, then $\alpha+\beta+\gamma$ is equal to:

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Three Dimensional Geometry

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

Δ PQR

, then

α + β + γ

is equal to:

(A)4
(B)5
(C)6✓ Correct
(D)8

Answer: (C)

Solution

A general point of the line is

(8 λ - 3, 2 λ + 4, 2 λ - 1)

. Its squared distance from

R (1, 2, 3)

(8 λ - 4)^{2} + (2 λ + 2)^{2} + (2 λ - 4)^{2} = 72 λ^{2} - 72 λ + 36

; setting this to

36

gives

λ (λ - 1) = 0

, so

λ = 0, 1

and

P (- 3, 4, - 1)

Q (5, 6, 1)

. The centroid of

△ P QR

(\frac{- 3 + 5 + 1}{3}, \frac{4 + 6 + 2}{3}, \frac{- 1 + 1 + 3}{3}) = (1, 4, 1)

, so

α + β + γ = 6

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Single correct option

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