JEE Main 2026 Apr 5 Shift 2, Mathematics Q15: Lines in Space

JEE Main2026Apr 5, Shift 2Mathematics

If the distance of the point $(a, 2, 5)$ from the image of the point $(1, 2, 7)$ in the line $\frac{x}{1} = \frac{y - 1}{1} = \frac{z - 2}{2}$ is 4, then the sum of all possible values of $a$ is equal to:

A
11
B
9
C
6
D
4

Solution

Let the image be $Q (α, β, γ)$ . The midpoint $M$ of $P Q$ lies on the line, and $P Q$ is perpendicular to the direction $(1, 1, 2)$ .

Perpendicularity: $(α - 1) + (β - 2) + 2 (γ - 7) = 0 \Rightarrow α + β + 2 γ = 17$ .

Midpoint on line: $\frac{( α + 1 ) /2}{1} = \frac{( β + 2 ) /2 - 1}{1} = \frac{( γ + 7 ) /2 - 2}{2}$ .

Simplifying gives $α + 1 = β$ and $2 β = γ + 3$ . Solving these with the perpendicularity condition: $α = 3$ , $β = 4$ , $γ = 5$ .

Distance from $(a, 2, 5)$ to $(3, 4, 5)$ : $(a - 3)^{2} + 4 = 4 \Rightarrow (a - 3)^{2} = 12$ , so $a^{2} - 6 a - 3 = 0$ .

Sum of roots $= 6$ .

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JEE Main 2026 Apr 5 Shift 2, Mathematics Q15: Lines in Space

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