Let \(\mathrm{L}_1: \frac{x+2}{5}=\frac{y-3}{2}=\frac{z-6}{1}\) and \(\mathrm{L}_2: \frac{x-3}{4}=\frac{y+2}{3}=\frac{z-3}{5}\) be the given lines, 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 \(\bar{b}_1=5 \hat{i}+2 \hat{j}+\hat{k}\) and \(\bar{b}_2=4 \hat{i}+3 \hat{j}+5 \hat{k}\) respectively.
\(\therefore \quad\) 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, \(\overline{\mathrm{b}}_1 \times \overline{\mathrm{b}}_2=\left|\begin{array}{lll}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 5 & 2 & 1 \\ 4 & 3 & 5\end{array}\right|=7 \hat{\mathrm{i}}-21 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}\)
\(\begin{aligned}
\therefore \quad \hat{\mathrm{n}} & =\frac{7 \hat{\mathrm{i}}-21 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}}{\sqrt{539}} \\
& =\frac{7(\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+\hat{\mathrm{k}})}{7 \sqrt{11}}=\frac{\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{11}}
\end{aligned}\)
[Note: The answer of the question is not mentioned as an option.]