The acute angle between the planes $$P_1$$ and $$P_2$$, when $$P_1$$ and $$P_2$$ are the planes passing through the intersection of the planes $$5x + 8y + 13z - 29 = 0$$ and $$8x - 7y + z - 20 = 0$$ and the points $$(2, 1, 3)$$ and $$(0, 1, 2)$$, respectively, is

The two given planes are:

$$P_1': 5x + 8y + 13z - 29 = 0 \quad \text{and} \quad P_2': 8x - 7y + z - 20 = 0$$

Any plane through their intersection can be written as:

$$(5x + 8y + 13z - 29) + \lambda(8x - 7y + z - 20) = 0$$

$$(5 + 8\lambda)x + (8 - 7\lambda)y + (13 + \lambda)z - (29 + 20\lambda) = 0$$

Finding $$P_1$$ (passes through $$(2, 1, 3)$$):

$$(5+8\lambda)(2) + (8-7\lambda)(1) + (13+\lambda)(3) = 29 + 20\lambda$$

$$10 + 16\lambda + 8 - 7\lambda + 39 + 3\lambda = 29 + 20\lambda$$

$$57 + 12\lambda = 29 + 20\lambda$$

$$28 = 8\lambda \implies \lambda = \frac{7}{2}$$

$$P_1: (5+28)x + (8-\frac{49}{2})y + (13+\frac{7}{2})z = 29 + 70$$

Multiplying through by 2: $$66x - 33y + 33z = 198$$

Normal: $$(66, -33, 33)$$ or $$(2, -1, 1)$$.

Finding $$P_2$$ (passes through $$(0, 1, 2)$$):

$$(5+8\mu)(0) + (8-7\mu)(1) + (13+\mu)(2) = 29 + 20\mu$$

$$8 - 7\mu + 26 + 2\mu = 29 + 20\mu$$

$$34 - 5\mu = 29 + 20\mu$$

$$5 = 25\mu \implies \mu = \frac{1}{5}$$

Normal of $$P_2$$: $$(5 + \frac{8}{5}, 8 - \frac{7}{5}, 13 + \frac{1}{5}) = (\frac{33}{5}, \frac{33}{5}, \frac{66}{5})$$

Simplified: $$(1, 1, 2)$$.

The angle between the planes with normals $$\vec{n_1} = (2, -1, 1)$$ and $$\vec{n_2} = (1, 1, 2)$$:

$$\cos\theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}||\vec{n_2}|} = \frac{|2 - 1 + 2|}{\sqrt{6} \cdot \sqrt{6}} = \frac{3}{6} = \frac{1}{2}$$

$$\theta = \frac{\pi}{3}$$

Hence the correct answer is Option A: $$\dfrac{\pi}{3}$$.