JEE Main 2026 — 5 April, Morning Shift — Question 38

Equate the two line equations: $(1 + a\lambda)\hat{i} + (1 - \lambda)\hat{j} + (-1)\hat{k} = (4 + 2\mu)\hat{i} + (0)\hat{j} + (-1 + a\mu)\hat{k}$

Compare coefficients: $1 + a\lambda = 4 + 2\mu$ $1 - \lambda = 0 \Rightarrow \lambda = 1$ $-1 = -1 + a\mu \Rightarrow a\mu = 0$

Since $a eq 0$, $\mu = 0$.

Substitute $\lambda = 1, \mu = 0$ into first equation: $a(1) = 3 + 2(0) \Rightarrow a = 3$

Point of intersection: $\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + 1(3\hat{i} - \hat{j}) = 4\hat{i} + 0\hat{j} - \hat{k}$

Distance from origin: $\sqrt{4^2 + 0^2 + (-1)^2} = \sqrt{16 + 0 + 1} = \sqrt{17}$

Square of distance: $(\sqrt{17})^2 = 17$