The shortest distance is given by,
= $\left|\frac{\left({\mathit{A}}_2-{\mathit{A}}_1\right).\left({\mathit{B}}_1\times {\mathit{B}}_2\right)}{\left|{\mathit{B}}_1\times {\mathit{B}}_2\right|}\right|$
${\mathit{A}}_2-{\mathit{A}}_1$ = $\left(\hat{\mathit{i}}-\hat{\mathit{j}}-2\hat{\mathit{k}}\right)$ - $\left(4\hat{\mathit{i}}-\hat{\mathit{j}}\right)$ = - $3\hat{\mathit{i}}-2\hat{\mathit{k}}$
${\mathit{B}}_1\times {\mathit{B}}_2$ = $\left|\begin{array}{ccc}\hat{\mathit{i}}& \hat{\mathit{j}}& \hat{\mathit{k}}\\ 1& 2& -3\\ 2& 4& -5\end{array}\right|$ = $2\hat{\mathit{i}}-\hat{\mathit{j}}$
$\left({\mathit{A}}_2-{\mathit{A}}_1\right)$ . $\left({\mathit{B}}_1\times {\mathit{B}}_2\right)$ = $\left(-3\hat{\mathit{i}}-2\hat{\mathit{k}}\right)$ . $\left(2\hat{\mathit{i}}-\hat{\mathit{j}}\right)$ = - 6
$\left|{\mathit{B}}_1\times {\mathit{B}}_2\right|$ = $\sqrt{2^2+1^2}$ = $\sqrt{5}$
so shortest distance between two lines = $\left|\frac{-6}{\sqrt{5}}\right|$ = $\frac{6}{\sqrt{5}}$ units