JEE Main 2026 Apr 5 Shift 2, Mathematics Q14: Lines in Space

JEE Main2026Apr 5, Shift 2Mathematics

Let a triangle $P QR$ 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 $△ P QR$ , then $α + β + γ$ is equal to:

Solution

Parametrise the line as $(8 λ - 3, 2 λ + 4, 2 λ - 1)$ . Distance squared from $R$ :

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

Dividing through by 4: $(4 λ - 2)^{2} + (λ + 1)^{2} + (λ - 2)^{2} = 9$ .

Expanding: $18 λ^{2} - 18 λ + 9 = 9 \Rightarrow λ (λ - 1) = 0$ , giving $λ = 0$ or $λ = 1$ .

So $P = (- 3, 4, - 1)$ , $Q = (5, 6, 1)$ , $R = (1, 2, 3)$ . Centroid $G = (\frac{- 3 + 5 + 1}{3}, \frac{4 + 6 + 2}{3}, \frac{- 1 + 1 + 3}{3}) = (1, 4, 1)$ .

$∴ α + β + γ = 6$ .

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JEE Main 2026 Apr 5 Shift 2, Mathematics Q14: Lines in Space

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