The user wants me to find the distance of a point from a plane. The plane is defined by a point it passes through and its normal vector, which is perpendicular to two given lines.

Step-by-step Solution:

1. Identify the direction vectors of the given lines. The equation of the first line is ${L_1}:{{x + 1} \over 3} = {{y + 2} \over 1} = {{z + 1} \over 2}$. The direction vector of ${L_1}$ is ${\vec d_1} = 3\hat i + 1\hat j + 2\hat k$.

   The equation of the second line is ${L_2}:{{x - 2} \over 1} = {{y - 2} \over 2} = {{z - 3} \over 3}$. The direction vector of ${L_2}$ is ${\vec d_2} = 1\hat i + 2\hat j + 3\hat k$.

2. Find the normal vector to the plane. The problem states that the normal to the plane, let's call it ${\vec n}$, is perpendicular to both lines ${L_1}$ and ${L_2}$. This means ${\vec n}$ is perpendicular to both ${\vec d_1}$ and ${\vec d_2}$. We can find ${\vec n}$ by taking the cross product of ${\vec d_1}$ and ${\vec d_2}$. ${\vec n} = {\vec d_1} \times {\vec d_2} = \begin{vmatrix} \hat i & \hat j & \hat k \\ 3 & 1 & 2 \\ 1 & 2 & 3 \end{vmatrix}$ Expanding the determinant: ${\vec n} = \hat i(1 \cdot 3 - 2 \cdot 2) - \hat j(3 \cdot 3 - 1 \cdot 2) + \hat k(3 \cdot 2 - 1 \cdot 1)$ ${\vec n} = \hat i(3 - 4) - \hat j(9 - 2) + \hat k(6 - 1)$ ${\vec n} = -1\hat i - 7\hat j + 5\hat k$ The direction ratios of the normal are $\langle -1, -7, 5 \rangle$.

3. Find the equation of the plane. The plane passes through the point $P(-1, -2, -1)$. The equation of a plane passing through a point $(x_0, y_0, z_0)$ with a normal vector with direction ratios $\langle A, B, C \rangle$ is given by $A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$.

   Using the point $P(-1, -2, -1)$ and the direction ratios of the normal $\langle -1, -7, 5 \rangle$: $-1(x - (-1)) - 7(y - (-2)) + 5(z - (-1)) = 0$ $-1(x + 1) - 7(y + 2) + 5(z + 1) = 0$ $-x - 1 - 7y - 14 + 5z + 5 = 0$ $-x - 7y + 5z - 10 = 0$ Multiplying the entire equation by $-1$ to make the coefficient of $x$ positive: $x + 7y - 5z + 10 = 0$

4. Calculate the distance of the point (1, 1, 1) from the plane. The distance of a point $(x_1, y_1, z_1)$ from a plane $Ax + By + Cz + D = 0$ is given by the formula: $d = {{|Ax_1 + By_1 + Cz_1 + D|} \over {\sqrt{A^2 + B^2 + C^2}}}$ The point is $Q(1, 1, 1)$ and the plane is $x + 7y - 5z + 10 = 0$. Here, $A=1$, $B=7$, $C=-5$, $D=10$, and $(x_1, y_1, z_1) = (1, 1, 1)$.

   $d = {{|1(1) + 7(1) - 5(1) + 10|} \over {\sqrt{1^2 + 7^2 + (-5)^2}}}$ $d = {{|1 + 7 - 5 + 10|} \over {\sqrt{1 + 49 + 25}}}$ $d = {{|13|} \over {\sqrt{75}}}$ $d = {{13} \over {\sqrt{75}}}$

5. Conclusion. The calculated distance is ${13 \over {\sqrt{75}}}$. Comparing this with the given options, it matches option C.

   A: ${2 \over {\sqrt {75} }}$ B: ${7 \over {\sqrt {75} }}$ C: ${13 \over {\sqrt {75} }}$ D: ${23 \over {\sqrt {75} }}$

The correct option is C.