Let the points on the plane $$P$$ be equidistant from the points $$(-4, 2, 1)$$ and $$(2, -2, 3)$$. Then the acute angle between the plane $$P$$ and the plane $$2x + y + 3z = 1$$ is

The plane $$P$$ is the locus of points equidistant from $$A(-4, 2, 1)$$ and $$B(2, -2, 3)$$. This is the perpendicular bisector plane of segment $$AB$$.

The midpoint of $$AB$$ is: $$M = \left(\frac{-4+2}{2}, \frac{2-2}{2}, \frac{1+3}{2}\right) = (-1, 0, 2)$$

The direction vector $$\vec{AB} = (2-(-4), -2-2, 3-1) = (6, -4, 2)$$.

The normal to plane $$P$$ is $$\vec{n_1} = (6, -4, 2)$$, which simplifies to $$(3, -2, 1)$$.

The equation of plane $$P$$ is: $$3(x+1) - 2(y-0) + 1(z-2) = 0$$

$$3x + 3 - 2y + z - 2 = 0$$

$$3x - 2y + z + 1 = 0$$

The second plane is $$Q: 2x + y + 3z = 1$$ with normal $$\vec{n_2} = (2, 1, 3)$$.

The angle between the planes is given by:

$$\cos\alpha = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}||\vec{n_2}|}$$

$$\vec{n_1} \cdot \vec{n_2} = 3(2) + (-2)(1) + 1(3) = 6 - 2 + 3 = 7$$

$$|\vec{n_1}| = \sqrt{9 + 4 + 1} = \sqrt{14}$$

$$|\vec{n_2}| = \sqrt{4 + 1 + 9} = \sqrt{14}$$

$$\cos\alpha = \frac{7}{\sqrt{14} \cdot \sqrt{14}} = \frac{7}{14} = \frac{1}{2}$$

Therefore $$\alpha = \frac{\pi}{3}$$.

The correct answer is Option C.