The distance of the point $$P(4, 6, -2)$$ from the line passing through the point $$(-3, 2, 3)$$ and parallel to a line with direction ratios $$3, 3, -1$$ is equal to:

We are given point $$P(4, 6, -2)$$ and a line passing through $$A(-3, 2, 3)$$ with direction ratios $$(3, 3, -1)$$, so $$\vec{AP} = (4-(-3), 6-2, -2-3) = (7, 4, -5).$$

The direction vector of the line is $$\vec{d} = (3, 3, -1)$$, and the cross product of $$\vec{AP}$$ with $$\vec{d}$$ is given by:

$$\vec{AP} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & 4 & -5 \\ 3 & 3 & -1 \end{vmatrix} = \hat{i}(4 \cdot (-1) - (-5) \cdot 3) - \hat{j}(7 \cdot (-1) - (-5) \cdot 3) + \hat{k}(7 \cdot 3 - 4 \cdot 3)$$

$$= \hat{i}(-4 + 15) - \hat{j}(-7 + 15) + \hat{k}(21 - 12) = (11, -8, 9)$$

The magnitude of this cross product is $$|\vec{AP} \times \vec{d}| = \sqrt{121 + 64 + 81} = \sqrt{266}$$ while the magnitude of $$\vec{d}$$ is $$|\vec{d}| = \sqrt{9 + 9 + 1} = \sqrt{19}$$. It follows that the distance from the point to the line is $$\frac{|\vec{AP} \times \vec{d}|}{|\vec{d}|} = \frac{\sqrt{266}}{\sqrt{19}} = \sqrt{\frac{266}{19}} = \sqrt{14}.$$

The correct answer is Option D: $$\sqrt{14}$$.