Let \(\mathrm{L}_1: \frac{x+1}{3}=\frac{y+2}{1}=\frac{\mathrm{z}+1}{2}\) and
\(\mathrm{L}_2: \frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{3}\)
be two given lines. Then the unit vector perpendicular to \(L_1\) and \(L_2\) is

Vector perpendicular to \(\mathrm{L}_1\) and \(\mathrm{L}_2\) is \(\mathrm{L}_1 \times \mathrm{L}_2\)
\(\begin{aligned}
\therefore \mathrm{L}_1 \times \mathrm{L}_2=\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{k} \\
3 & 1 & 2 \\
1 & 2 & 3
\end{array}\right|=-\hat{\mathrm{i}}-7 \hat{\mathbf{j}}+5 \hat{\mathrm{k}}\end{aligned}\)
\(\therefore \text { Required unit vector }=\frac{-\hat{\mathrm{i}}-7 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}}{\sqrt{(-1)^2+(-7)^2+(5)^2}}\)
\(=\frac{-\hat{\mathrm{i}}-7 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}}{\sqrt{75}}=\frac{-\hat{\mathrm{i}}-7 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}}{5 \sqrt{3}}\)