Consider the lines
\(\mathrm{L}_1: \frac{x+1}{3}=\frac{y+2}{1}=\frac{\mathrm{z}+1}{2}\) \(\mathrm{L}_2: \frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{3}\)
then the unit vector perpendicular to both \(\mathrm{L}_1\) and \(\mathrm{L}_2\) is

Lines \(L_1\) and \(L_2\) are parallel to the vectors
\(
\overline{\mathrm{b}}_1=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}} \text { and } \overline{\mathrm{b}}_2=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}
\)
respectively.
\(\therefore\) The unit vector perpendicular to both \(\mathrm{L}_1\) and \(\mathrm{L}_2\) is \(\hat{\mathrm{n}}=\frac{\overline{\mathrm{b}}_1 \times \overline{\mathrm{b}}_2}{\left|\overline{\mathrm{b}}_1 \times \overline{\mathrm{b}}_2\right|}\)
Now, \(\bar{b}_1 \times \bar{b}_2=\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ 1 & 2 & 3\end{array}\right|=-\hat{\mathrm{i}}-7 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\)
\(\therefore \hat{\mathrm{n}}=\frac{1}{5 \sqrt{3}}(-\hat{\mathrm{i}}-7 \hat{\mathrm{j}}+5 \hat{\mathrm{k}})\)