Let $$P(3, 2, 3), Q(4, 6, 2)$$ and $$R(7, 3, 2)$$ be the vertices of $$\triangle PQR$$. Then, the angle $$\angle QPR$$ is

Given points $$P(3, 2, 3)$$, $$Q(4, 6, 2)$$, and $$R(7, 3, 2)$$. We need to find angle $$\angle QPR$$.

Find the direction vectors from P: $$\vec{PQ} = Q - P = (4-3, 6-2, 2-3) = (1, 4, -1)$$

$$\vec{PR} = R - P = (7-3, 3-2, 2-3) = (4, 1, -1)$$

Compute the dot product and magnitudes: $$\vec{PQ} \cdot \vec{PR} = (1)(4) + (4)(1) + (-1)(-1) = 4 + 4 + 1 = 9$$

$$|\vec{PQ}| = \sqrt{1 + 16 + 1} = \sqrt{18} = 3\sqrt{2}$$

$$|\vec{PR}| = \sqrt{16 + 1 + 1} = \sqrt{18} = 3\sqrt{2}$$

Apply the angle formula: $$ \cos(\angle QPR) = \frac{\vec{PQ} \cdot \vec{PR}}{|\vec{PQ}||\vec{PR}|} = \frac{9}{3\sqrt{2} \cdot 3\sqrt{2}} = \frac{9}{18} = \frac{1}{2} $$

$$ \angle QPR = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} $$

The correct answer is Option (4): $$\frac{\pi}{3}$$.