Let \(\mathrm{L}_1\) \(\frac{x+1}{3}=\frac{y+2}{2}=\frac{z+1}{1}\) and \(L_2: \frac{x-2}{2}=\frac{y+2}{1}=\frac{z-3}{3}\) be the given lines. Then the unit vector perpendicular to \(L_1\) and \(L_2\) is

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