Here, ${\overrightarrow{a}}_1=\hat{i}+\hat{j} \& {\overrightarrow{b}}_1=2\hat{i}-\hat{j}+\hat{k}$ and ${\overrightarrow{a}}_2=2\hat{i}+\hat{j}-\hat{k} \& {\overrightarrow{b}}_2=3\hat{i}-5\hat{j}+2\hat{k}$.

$\Rightarrow {\overrightarrow{a}}_2-{\overrightarrow{a}}_2=\hat{i}-\hat{k} \& {\overrightarrow{b}}_1\times {\overrightarrow{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -1& 1\\ 3& -5& 2\end{array}\right|=3\hat{i}-\hat{j}-7\hat{k}$

The shortest distance between lines $L_1$ and $L_2 =\left|\frac{\left({\overrightarrow{a}}_2-{\overrightarrow{a}}_2\right)·\left({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2\right)}{ {\overrightarrow{b}}_1\times {\overrightarrow{b}}_2}\right|$

$=\left|\frac{\left(\hat{i}-\hat{k}\right)·\left(3\hat{i}-\hat{j}-7\hat{k}\right)}{3\hat{i}-\hat{j}-7\hat{k}}\right|$

$=\frac{10}{\sqrt{59}}$