Find Distance of Point from Plane Normal to Two Lines

To find the distance of the point $(1, 1, 1)$ from the given plane, we first need to determine the equation of the plane.

Step 1: Find the direction vector normal to the plane

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

The direction vectors of lines $L_1$ and $L_2$ are:
$\vec{b}_1 = 3\hat{i} + \hat{j} + 2\hat{k}$
$\vec{b}_2 = \hat{i} + 2\hat{j} + 3\hat{k}$

Since the normal to the plane, $\vec{n}$, is perpendicular to both $L_1$ and $L_2$, it is given by the cross product of $\vec{b}_1$ and $\vec{b}_2$:

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

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

Step 2: Find the equation of the plane

The plane passes through the point $(-1, -2, -1)$ and has a normal vector $\vec{n} = -\hat{i} - 7\hat{j} + 5\hat{k}$.

The equation of a plane passing through $(x_0, y_0, z_0)$ with normal vector $a\hat{i} + b\hat{j} + c\hat{k}$ is:
$a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$

Substituting the values:
$-1(x - (-1)) - 7(y - (-2)) + 5(z - (-1)) = 0$
$-1(x + 1) - 7(y + 2) + 5(z + 1) = 0$
$-x - 1 - 7y - 14 + 5z + 5 = 0$
$-x - 7y + 5z - 10 = 0$

Multiplying by $-1$, we get the standard equation of the plane:
$x + 7y - 5z + 10 = 0$

Step 3: Calculate the distance of the point $(1, 1, 1)$ from the plane

The perpendicular distance $d$ of a point $(x_1, y_1, z_1)$ from a plane $Ax + By + Cz + D = 0$ is given by:
$d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$

Substituting $(x_1, y_1, z_1) = (1, 1, 1)$ and the plane parameters $A = 1, B = 7, C = -5, D = 10$:
$d = \frac{|1(1) + 7(1) - 5(1) + 10|}{\sqrt{1^2 + 7^2 + (-5)^2}}$
$d = \frac{|1 + 7 - 5 + 10|}{\sqrt{1 + 49 + 25}}$
$d = \frac{13}{\sqrt{75}}$

Thus, the correct option is C.