We have lines $\overline{\mathrm{r}}=(2 \hat{\mathrm{i}}-\hat{\mathrm{j}})+\lambda(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-3 \hat{\mathrm{k}})$
and $\overline{\mathrm{r}}=(\hat{\mathrm{r}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})+\mu(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-5 \hat{\mathrm{k}})$
Let $\bar{a}=2 \hat{i}-\hat{j}$ and $\bar{b}=\hat{i}-\hat{j}+2 \hat{k}$
$$
\therefore \overline{\mathrm{AB}}=-\hat{\mathrm{i}}+2 \hat{\mathrm{k}}
$$
Vector perpendicular to given lines is
$$
\overline{\mathrm{n}}=\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
2 & 1 & -3 \\
2 & 1 & -5
\end{array}\right|=\hat{\mathrm{i}}(-5+3)-\hat{\mathrm{j}}(-10+6)+\hat{\mathrm{k}}(2-2)=-2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}
$$
Shortest distance between given lines
$$
=\overline{\mathrm{AB}} \cdot \frac{\overline{\mathrm{n}}}{|\overline{\mathrm{n}}|}=\frac{(-\hat{\mathrm{i}}+2 \hat{\mathrm{k}}) \cdot(-2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}})}{\sqrt{(-2)^2+(4)^2}}=\frac{2}{\sqrt{20}}=\frac{1}{\sqrt{5}}
$$