QJEEM25Ma1ae0: JEE Main 2025 Mathematics - Class 12 JEEM25Ma1ae0 Solution: The square of the distance of the point from the line in the…

$\begin{aligned}& L=\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7} \\& P Q=\frac{x-\frac{15}{7}}{1}=\frac{y-\frac{32}{7}}{4}=\frac{z-7}{7}=\lambda\end{aligned}$

$\begin{aligned}&\Rightarrow \mathrm{Q}\left(\lambda+\frac{15}{7}, 4 \lambda+\frac{32}{7}, 7 \lambda+7\right)\\&\text { Since } \mathrm{Q} \text { lies on line } \mathrm{L}\\&\begin{aligned}& \text { So, } \frac{\lambda+\frac{15}{7}+1}{3}=\frac{7 \lambda+7+5}{7} \\& \Rightarrow 7 \lambda+22=21 \lambda+36 \\& \Rightarrow \lambda=-1 \\& \therefore \text { Point } Q\left(\frac{8}{7}, \frac{4}{7}, 0\right) \\& \mathrm{PQ}=\sqrt{\left(\frac{15}{7}-\frac{8}{7}\right)^2+\left(\frac{32}{7}-\frac{4}{7}\right)^2+(7-0)} \\& \mathrm{PQ}=\sqrt{66} \\& \Rightarrow(\mathrm{PQ})^2=66\end{aligned}\end{aligned}$