JEE Advanced
Question 30 — JEE Main 28 Jan 2026 S Ii
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Solution
line $\mathrm{L}_{1}: \frac{\mathrm{x}-4}{2}=\frac{\mathrm{y}-0}{0}=\frac{\mathrm{z}+1}{3}=\lambda$
line $\mathrm{L}_{2}: \frac{\mathrm{x}-43}{3}=\frac{\mathrm{y}-\alpha}{-1}=\frac{\mathrm{z}-\beta}{0}=\mu$
$\mathrm{Q}(2 \lambda+4,0,3 \lambda-1) \equiv \mathrm{Q}(3 \mu+43,-\mu+\alpha, \beta)$\[ 2 \lambda+4=3 \mu+43 \Rightarrow 2 \lambda=3 \mu+39 \]\[ 0=-\mu+\alpha=\alpha=\mu \]
\[ 3 \lambda-1=\beta \]\[ \mathrm{Q}(3 \alpha+43,0, \beta) \mathrm{P}(43, \alpha, \beta) \]\[ \begin{aligned} & P Q=13 \sqrt{10} \Rightarrow \sqrt{9 \alpha^{2}+\alpha^{2}+0}=13 \sqrt{10} \\ & \Rightarrow|\alpha| \sqrt{10}=13 \sqrt{10} \Rightarrow \alpha= \pm 13 \end{aligned} \]Case I: $\alpha=13, \mu=13$, given $\beta<0$
From (1), $2 \lambda=3 \mu+39=78 \Rightarrow \lambda=39$
From (2), $\beta=3 \lambda-1=3 \times 39-1>0$ (rejected)
Case II : $\alpha=-13, \mu=-13$
From (1), $2 \lambda=-39+39 \Rightarrow \lambda=0$
From (3), $\beta=3 \lambda-1=-1$
Now $\alpha^{2}+\beta^{2}=169+1=170$ Answer(170)
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