Let $$PQR$$ be a triangle with $$R(-1, 4, 2)$$. Suppose $$M(2, 1, 2)$$ is the mid point of $$PQ$$. The distance of the centroid of $$\Delta PQR$$ from the point of intersection of the line $$\frac{x-2}{0} = \frac{y}{2} = \frac{z+3}{-1}$$ and $$\frac{x-1}{1} = \frac{y+3}{-3} = \frac{z+1}{1}$$ is

Given triangle PQR with vertex R(-1, 4, 2) and M(2, 1, 2) as the midpoint of PQ. The centroid of triangle PQR needs to be found, and then its distance from the intersection point of the lines $$\frac{x-2}{0} = \frac{y}{2} = \frac{z+3}{-1}$$ and $$\frac{x-1}{1} = \frac{y+3}{-3} = \frac{z+1}{1}$$.

Let the coordinates of P and Q be $$P(x_1, y_1, z_1)$$ and $$Q(x_2, y_2, z_2)$$. Since M(2, 1, 2) is the midpoint of PQ:

$$\frac{x_1 + x_2}{2} = 2 \implies x_1 + x_2 = 4$$

$$\frac{y_1 + y_2}{2} = 1 \implies y_1 + y_2 = 2$$

$$\frac{z_1 + z_2}{2} = 2 \implies z_1 + z_2 = 4$$

The centroid G of triangle PQR is the average of the coordinates of P, Q, and R(-1, 4, 2):

$$G_x = \frac{x_1 + x_2 + (-1)}{3} = \frac{4 + (-1)}{3} = \frac{3}{3} = 1$$

$$G_y = \frac{y_1 + y_2 + 4}{3} = \frac{2 + 4}{3} = \frac{6}{3} = 2$$

$$G_z = \frac{z_1 + z_2 + 2}{3} = \frac{4 + 2}{3} = \frac{6}{3} = 2$$

Thus, the centroid G is at (1, 2, 2).

Next, find the intersection point of the two lines. The first line is $$\frac{x-2}{0} = \frac{y}{2} = \frac{z+3}{-1}$$. Since the denominator of x is 0, x is constant: x = 2. Let the parameter be λ:

$$\frac{y}{2} = \lambda \implies y = 2\lambda$$

$$\frac{z+3}{-1} = \lambda \implies z = - \lambda - 3$$

So parametric equations are: x = 2, y = 2λ, z = -λ - 3.

The second line is $$\frac{x-1}{1} = \frac{y+3}{-3} = \frac{z+1}{1}$$. Let the parameter be μ:

$$\frac{x-1}{1} = \mu \implies x = 1 + \mu$$

$$\frac{y+3}{-3} = \mu \implies y = -3 - 3\mu$$

$$\frac{z+1}{1} = \mu \implies z = -1 + \mu$$

So parametric equations are: x = 1 + μ, y = -3 - 3μ, z = -1 + μ.

Set the coordinates equal to find intersection:

From x-coordinates: 2 = 1 + μ ⟹ μ = 1

Substitute μ = 1 into z-coordinate: z = -1 + 1 = 0

Set z from first line: -λ - 3 = 0 ⟹ λ = -3

Substitute λ = -3 into y-coordinate: y = 2(-3) = -6

Verify with second line y-coordinate: y = -3 - 3(1) = -6, which matches.

Thus, the intersection point S is (2, -6, 0).

The distance between G(1, 2, 2) and S(2, -6, 0) is:

$$d = \sqrt{(2-1)^2 + (-6-2)^2 + (0-2)^2} = \sqrt{(1)^2 + (-8)^2 + (-2)^2} = \sqrt{1 + 64 + 4} = \sqrt{69}$$

Therefore, the distance is $$\sqrt{69}$$.