JEE Main 2025 Jan 28 Shift 2, Mathematics Q8: Lines in Space

Let $P=(15/7,32/7,7)$ and $Q$ be the foot of the line through $P$ in direction $\hat{i}+4\hat{j}+7\hat{k}$ meeting the given line $L:\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$.

Parametrise $Q=P+\lambda (1,4,7)=\left(\frac{15}{7}+\lambda ,\frac{32}{7}+4\lambda ,7+7\lambda \right)$. Imposing the condition that $Q$ lies on $L$ (equating the first and third ratios):

$\frac{(15/7+\lambda )+1}{3}=\frac{(7+7\lambda )+5}{7}\Rightarrow 7\lambda +22=21\lambda +36\Rightarrow \lambda =-1$.

So $Q=(8/7,4/7,0)$, and

$P{Q}^2=(15/7-8/7{)}^2+(32/7-4/7{)}^2+7^2=1+16+49=66$.