Consider the lines L _ 1 and L _ 2 given by L _ 1 : (x - 1)/(2) =...

Given,

The lines ${L}_{1} and${L}_{2}$given by</p><p>${L}_{1}: \dfrac{x - 1}{2} = \dfrac{y - 3}{1} = \dfrac{z - 2}{2}$</p><p>${L}_{2}: \dfrac{x - 2}{1} = \dfrac{y - 2}{2} = \dfrac{z - 3}{3}$</p><p>Now any point on line${L}_{1}$is given by,</p><p>$P = \left(2\lambda + 1, \lambda + 3, 2\lambda + 2\right)$</p><p>And any point on line${L}_{2}$is given by,</p><p>$Q = \left(\mu + 2, 2\mu + 2, 3\mu + 3\right)$</p><p>Now equation of line${L}_{3}$which passes through$Q$is given by,</p><p>$\dfrac{x - \left(\mu + 1\right)}{1} = \dfrac{y - \left(2\mu + 2\right)}{- 1} = \dfrac{z - \left(3\mu + 3\right)}{- 2}$</p><p>Now it also passes through$P$, so we get,</p><p>$⇒ \dfrac{2\lambda - \mu - 1}{1} = \dfrac{\lambda - 2\mu + 1}{- 1} = \dfrac{2\lambda - 3\mu - 1}{- 2}$</p><p>Now on solving above equations we get,$\lambda = \mu = 3$</p><p>Hence,$$P\left(7, 6, 8\right)$ and $Q\left(5, 8, 12\right)$

So, the distance $PQ = 2\sqrt{6}$