Let $\mathrm{L}_1: \frac{x+2}{5}=\frac{y-3}{2}=\frac{z-6}{1}$ and $\mathrm{L}_2:…
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.]