JEE Main 2026 Apr 8 Shift 2, Mathematics Q23: Lines in Space

JEE Main2026Apr 8, Shift 2Mathematics

Let a line $L_{1}$ pass through the origin and be perpendicular to the lines

$L_{2} : r = (3 + t) \hat{i} + (2 t - 1) \hat{j} + (2 t + 4) \hat{k}$ and

$L_{3} : r = (3 + 2 s) \hat{i} + (3 + 2 s) \hat{j} + (2 + s) \hat{k}$ , $t, s \in R$ . If $(a, b, c)$ , $a \in Z$ , is the point on $L_{3}$ at a distance of $17$ from the point of intersection of $L_{1}$ and $L_{2}$ , then $(a + b + c)^{2}$ is equal to _______.

Your answer

Solution

Direction of $L_{1}$ : perpendicular to both $d_{2} = (1, 2, 2)$ and $d_{3} = (2, 2, 1)$ :

$d_{2} \times d_{3} = (2 \cdot 1 - 2 \cdot 2, 2 \cdot 2 - 1 \cdot 1, 1 \cdot 2 - 2 \cdot 2) = (- 2, 3, - 2)$

So $L_{1} : \frac{x}{2} = \frac{y}{- 3} = \frac{z}{2}$ (taking the opposite sign for convenience).

Intersection of $L_{1}$ and $L_{2}$ : A point on $L_{2}$ is $(3 + t, 2 t - 1, 2 t + 4)$ . For it to lie on $L_{1}$ (passing through origin with direction $(2, - 3, 2)$ ), the coordinates must be proportional. From $\frac{3 + t}{2} = \frac{2 t + 4}{2}$ we get $t = - 1$ , giving the point $(2, - 3, 2)$ .

Point on $L_{3}$ at distance $17$ : A generic point is $Q = (3 + 2 s, 3 + 2 s, 2 + s)$ . Distance squared from $(2, - 3, 2)$ :

$(2 s + 1)^{2} + (2 s + 6)^{2} + s^{2} = 17$

$4 s^{2} + 4 s + 1 + 4 s^{2} + 24 s + 36 + s^{2} = 17 ⟹ 9 s^{2} + 28 s + 20 = 0$

Roots: $s = - 2$ and $s = - \frac{10}{9}$ . For $a \in Z$ we need $3 + 2 s$ integer, so $s = - 2$ , giving $Q = (- 1, - 1, 0)$ .

$(a + b + c)^{2} = (- 2)^{2} = 4$ .

Concept behind this question

Lines in SpaceNotes, formulas and examples →

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2026 Apr 8 Shift 2, Mathematics Q23: Lines in Space

Concept behind this question

More previous year questions on Lines in Space

Practice more Mathematics