Mathematics - Vector Question with Solution | TestHub

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