The shortest distance between the lines $\vec{r} = (4\hat{i} - \hat{j})+λ(\hat{i} +2\hat{j}-3\hat{k})$

and, $\vec{r}=(\hat{i} -\hat{j}+2\hat{k})+\mu (2\hat{i} +4\hat{j}-5\hat{k})$ is

$\frac{6}{\sqrt{5}}$

We know that the shortest distance between the lines $\vec{r}= \vec{a_1} + λ\vec{b_1}$ and $\vec{r}= \vec{a_2} + \mu \vec{b_2}$ is given by 

$d = \begin{vmatrix} \frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}\end{vmatrix}$

Hence, $\vec{a_1}= 4 \hat{i} - \hat{j}, \vec{a_2}= \hat{i}-\hat{j} + 2\hat{k}$,

$\vec{b_1} = \hat{i} + 2\hat{j} - 3\hat{k} $ and $\vec{b_2} =2 \hat{i} + 4\hat{j} - 5\hat{k}$

$∴ \vec{a_2}- \vec{a_1}= -3\hat{i} + 0\hat{j} +2\hat{k}$

and $\vec{b_1}×\vec{b_2}= \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\1 & 2 & -3\\2 & 4 & -5\end{vmatrix}=2\hat{i} -\hat{j} +0\hat{k}$

$⇒(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})= ( -3\hat{i} + 0\hat{j} +2\hat{k}).(2\hat{i} -\hat{j} +0\hat{k})= - 6$

and , $|\vec{b_1}×\vec{b_2}|= \sqrt{4+1+0}= \sqrt{5}$

$∴ d = \begin{vmatrix} \frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}\end{vmatrix}= \left|\frac{-6}{\sqrt{5}}\right|= \frac{6}{\sqrt{5}}$