JEE Main 2025 — 28 January, Evening Shift — Question 8

$L=\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$

$P Q=\frac{x-\frac{15}{7}}{1}=\frac{y-\frac{32}{7}}{4}=\frac{z-7}{7}=\lambda$

$\Rightarrow \mathrm{Q}\left(\lambda+\frac{15}{7}, 4 \lambda+\frac{32}{7}, 7 \lambda+7\right)$

Since Q lies on line L

So, $\frac{\lambda+\frac{15}{7}+1}{3}=\frac{7 \lambda+7+5}{7}$

$\Rightarrow 7 \lambda+22=21 \lambda+36$

$\Rightarrow \lambda=-1$

$\therefore$ Point $\mathrm{Q}\left(\frac{8}{7}, \frac{4}{7}, 0\right)$

$P Q=\sqrt{\left(\frac{15}{7}-\frac{8}{7}\right)^{2}+\left(\frac{32}{7}-\frac{4}{7}\right)^{2}+(7-0)^{2}}$

$\mathrm{PQ}=\sqrt{66}$

$\Rightarrow(\mathrm{PQ})^{2}=66$