? $ \begin{aligned} & \mathrm{L}_1 \equiv \frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}-4}{5}=\frac{\mathrm{z}-2}{1}=\lambda \\ & \mathrm{P}(\lambda+2,5 \lambda+4, \lambda+2) \\ & \mathrm{L}_2 \equiv \frac{\mathrm{x}-3}{2}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{z}-3}{2} \\ & \mathrm{P}(2 \mu+3,3 \mu+2,2 \mu+3) \\ & \lambda+2=2 \mu+33 \mu+2=5 \lambda+4 \\ & \lambda=2 \mu+13 \mu=5 \lambda+2 \\ & 3 \mu=5(2 \mu+1)+2 \\ & 3 \mu=10 \mu+7 \\ & \mu=-1 \lambda=-1 \end{aligned} $
Both satisfies ( P )
$ \begin{aligned} & \mathrm{P}(1,-1,1) \\ & \mathrm{L}_3 \equiv \frac{\mathrm{x}}{1 / 4}=\frac{\mathrm{y}}{1 / 2}=\frac{\mathrm{z}}{1} \\ & \mathrm{~L}_3=\frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}}{4}=\mathrm{k} \end{aligned} $
Coordinates of Q(k, $2 \mathrm{k}, 4 \mathrm{k}$ )
DR's of $\mathrm{PQ}=\langle\mathrm{k}-1,2 \mathrm{k}+1,4 \mathrm{k}-1\rangle$
$\mathrm{PQ} \perp$ to $\mathrm{L}_3$
$ \begin{aligned} & (k-1)+2(2 k+1)+4(4 k-1)=0 \\ & k-1+4 k+2+16 k-4=0 \end{aligned} $
$ \begin{aligned} & \mathrm{k}=\frac{1}{7} \\ & \mathrm{Q}\left(\frac{1}{7}, \frac{2}{7}, \frac{4}{7}\right) \\ & \mathrm{PQ}=\sqrt{\left(1-\frac{1}{7}\right)^2+\left(-1-\frac{2}{7}\right)^2+\left(1-\frac{4}{7}\right)^2} \\ & =\sqrt{\frac{36}{49}+\frac{81}{49}+\frac{9}{49}}=\frac{\sqrt{126}}{7} \\ & \mathrm{PQ}=\frac{3 \sqrt{14}}{7} \end{aligned} $
Option-3 will satisfy