The distance of the point $$(-1, 2, 3)$$ from the plane $$\vec{r} \cdot (\hat{i} - 2\hat{j} + 3\hat{k}) = 10$$ parallel to the line of the shortest distance between the lines $$\vec{r} = (\hat{i} - \hat{j}) + \lambda(2\hat{i} + \hat{k})$$ and $$\vec{r} = (2\hat{i} - \hat{j}) + \mu(\hat{i} - \hat{j} + \hat{k})$$ is

We need to find the distance from $$(-1, 2, 3)$$ to the plane $$\vec{r} \cdot (\hat{i} - 2\hat{j} + 3\hat{k}) = 10$$, measured parallel to the line of shortest distance between two given lines.

Identify the plane equation. The plane equation in Cartesian form is: $$x - 2y + 3z = 10$$

Check if $$(-1, 2, 3)$$ lies on the plane: $$-1 - 4 + 9 = 4 \neq 10$$. The point is not on the plane.

Find the direction of shortest distance between the two lines. Line 1: $$\vec{r} = (\hat{i} - \hat{j}) + \lambda(2\hat{i} + \hat{k})$$, so direction $$\vec{d_1} = (2, 0, 1)$$

Line 2: $$\vec{r} = (2\hat{i} - \hat{j}) + \mu(\hat{i} - \hat{j} + \hat{k})$$, so direction $$\vec{d_2} = (1, -1, 1)$$

The direction of the line of shortest distance is $$\vec{d_1} \times \vec{d_2}$$:

$$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 1 \\ 1 & -1 & 1 \end{vmatrix}$$

$$= \hat{i}(0 \cdot 1 - 1 \cdot (-1)) - \hat{j}(2 \cdot 1 - 1 \cdot 1) + \hat{k}(2 \cdot (-1) - 0 \cdot 1)$$

$$= \hat{i}(0 + 1) - \hat{j}(2 - 1) + \hat{k}(-2 - 0) = (1, -1, -2)$$

Write the parametric line from $$(-1, 2, 3)$$ along $$(1, -1, -2)$$: $$(x, y, z) = (-1 + t, \; 2 - t, \; 3 - 2t)$$

Find the intersection with the plane $$x - 2y + 3z = 10$$: $$(-1 + t) - 2(2 - t) + 3(3 - 2t) = 10$$

$$-1 + t - 4 + 2t + 9 - 6t = 10$$

$$4 - 3t = 10$$

$$-3t = 6 \implies t = -2$$

Calculate the distance. The distance is $$|t|$$ times the magnitude of the direction vector:

$$|\vec{d}| = \sqrt{1^2 + (-1)^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$$

$$\text{Distance} = |{-2}| \times \sqrt{6} = 2\sqrt{6}$$

The correct answer is Option C: $$2\sqrt{6}$$.