A line with direction ratios $(2,7,-5)$ is drawn to intersect the lines $\frac{x-5}{3}=\frac{y-7}{-1

Question

A line with direction ratios $(2,7,-5)$ is drawn to intersect the lines $\frac{x-5}{3}=\frac{y-7}{-1}=\frac{z+2}{1}$ and $\frac{x+3}{-3}=\frac{y-3}{2}=\frac{z-6}{4}$ at P and Q respectively $(	ext { length of } P Q)^2$ is $\_\_\_\_$

Professor PC Thomas Classes · Answer

$\begin{aligned} & \mathrm{P}(3 \lambda+5-\lambda+7 \quad \lambda-2), \mathrm{Q}(-3 \mu+5 \quad 2 \mu+3 \quad 4 \mu+6) \ & 	ext { Drs of } \mathrm{PQ}(3 \lambda+3 \mu+8,-\lambda-2 \mu+4, \lambda-4 \mu-8) \ & \frac{3 \lambda+3 \mu+8}{2}=\frac{-\lambda-2 \mu+4}{7}=\frac{\lambda-4 \mu-8}{-5} \ & \Rightarrow \lambda=\mu=-1 \ & 	herefore P(2,8,-3) Q(0,1,2) \ & 	herefore P Q^2=4+49+25=78\end{aligned}$