JEE Main 2026 — Jan 22, Shift 2 — Mathematics solution

JEE Main2026Jan 22, Shift 2Mathematics

Let $L$ be the line $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z + 3}{6}$ and let $S$ be the set of all points $(a, b, c)$ on $L$ , whose distance from the line $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z - 9}{0}$ along the line $L$ is 7. Then $(a, b, c) \in S \sum (a + b + c)$ is equal to :

A
34
B
28
C
40
D
6

Solution

Let $M$ be the intersection of the two lines. Parameterising both and equating: $2 λ - 1 = 2 μ - 1$ , $3 λ - 1 = 3 μ - 1$ , $6 λ - 3 = 9 \Rightarrow λ = 2$ , giving $M = (3, 5, 9)$ .

A point on $L$ at parameter $K$ is $P = (2 K - 1, 3 K - 1, 6 K - 3)$ . Setting $∣ P M ∣ = 7$ :

$(2 K - 4)^{2} + (3 K - 6)^{2} + (6 K - 12)^{2} = 49 \Rightarrow 49 (K - 2)^{2} = 49 \Rightarrow K = 1$ or $3$ .

This gives points $(1, 2, 3)$ and $(5, 8, 15)$ . Sum of all coordinates $= (1 + 2 + 3) + (5 + 8 + 15) = 6 + 28 = 34$ .

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