Find Unit Vector Perpendicular to Two Given Lines

To find the unit vector perpendicular to both lines $L_1$ and $L_2$, we first determine the direction vectors of each line.

The given lines are:
$L_1 : \frac{x+1}{3} = \frac{y+2}{1} = \frac{z+1}{2}$
$L_2 : \frac{x-2}{1} = \frac{y+2}{2} = \frac{z-3}{3}$

From the denominators of the symmetric form of the lines, the direction vector of line $L_1$ is:
$\vec{b}_1 = 3\hat{i} + \hat{j} + 2\hat{k}$

And the direction vector of line $L_2$ is:
$\vec{b}_2 = \hat{i} + 2\hat{j} + 3\hat{k}$

A vector perpendicular to both $L_1$ and $L_2$ is given by the cross product of their direction vectors $\vec{b}_1$ and $\vec{b}_2$:

$$\vec{n} = \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ 1 & 2 & 3 \end{vmatrix}$$

Evaluating the determinant:
$\vec{n} = \hat{i}(1 \cdot 3 - 2 \cdot 2) - \hat{j}(3 \cdot 3 - 2 \cdot 1) + \hat{k}(3 \cdot 2 - 1 \cdot 1)$
$\vec{n} = \hat{i}(3 - 4) - \hat{j}(9 - 2) + \hat{k}(6 - 1)$
$\vec{n} = -\hat{i} - 7\hat{j} + 5\hat{k}$

Next, we find the magnitude of vector $\vec{n}$:
$|\vec{n}| = \sqrt{(-1)^2 + (-7)^2 + 5^2} = \sqrt{1 + 49 + 25} = \sqrt{75} = 5\sqrt{3}$

Thus, the unit vector perpendicular to both lines is:
$\hat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{-\hat{i} - 7\hat{j} + 5\hat{k}}{5\sqrt{3}}$

This corresponds to Option B.