Let ${\mathrm{L}}_1:\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{z}-3}{4}$ and ${\mathrm{L}}_2:\frac{\mathrm{x}-2}{3}=\frac{\mathrm{y}-4}{4}=\frac{\mathrm{z}-5}{5}$ be two lines. Then which of the following points lies on the line of the shortest distance between L₁ and L₂ ? 

$\mathrm{P}(2\lambda +1,3\lambda +2,4\lambda +3)$ on $L_1$

$\mathrm{Q}(3\mu +2,4\mu +4,5\mu +5)$ on $L_2$

Dr's of PQ=$3\mu -2\lambda +1,4\mu -3\lambda +2,5\mu -4\lambda +2$

PQ $\perp$ $L_1$

$\left(3\mu -2\lambda +1\right)2+\left(4\mu -3\lambda +2\right)3+\left(5\mu -4\lambda +2\right)4=0$

$38\mu -29\lambda +16=0 ...\left(1\right)$

PQ $\perp$ $L_2$

$\left(3\mu -2\lambda +1\right)3+\left(4\mu -3\lambda +2\right)4+\left(5\mu -4\lambda +2\right)5=0$

$50\mu -38\lambda +21=0 ...\left(2\right)$

Solving eq(1) and (2)

$\lambda =\frac{1}{3};\mu =\frac{-1}{6}$

$\therefore \mathrm{P}\left(\frac{5}{3},3,\frac{13}{3}\right)\mathrm{\&}\mathrm{Q}\left(\frac{3}{2},\frac{10}{3},\frac{25}{6}\right)$

Line PQ is

$\frac{x-\frac{5}{3}}{\frac{1}{6}}=\frac{y-3}{\frac{-1}{3}}=\frac{z-\frac{13}{3}}{\frac{1}{6}}$

$\frac{x-\frac{5}{3}}{1}=\frac{y-3}{-2}=\frac{z-\frac{13}{3}}{1}$

Point $\left(\frac{14}{3},-3,\frac{22}{3}\right)$ lies on the line PQ