Let the image of the point $$\left(\frac{5}{3}, \frac{5}{3}, \frac{8}{3}\right)$$ in the plane $$x - 2y + z - 2 = 0$$ be $$P$$. If the distance of the point $$Q(6, -2, \alpha)$$, $$\alpha > 0$$, from $$P$$ is $$13$$, then $$\alpha$$ is equal to _____.

We need to find the image of $$P_0 = \left(\frac{5}{3}, \frac{5}{3}, \frac{8}{3}\right)$$ in the plane $$x - 2y + z - 2 = 0$$ and then find $$\alpha$$.

Find the image $$P$$ using the reflection formula. For a plane $$ax + by + cz + d = 0$$, the image of a point is:

$$P' = P_0 - \frac{2(aP_{0x} + bP_{0y} + cP_{0z} + d)}{a^2 + b^2 + c^2} \cdot (a, b, c)$$

Here $$a = 1, b = -2, c = 1, d = -2$$.

$$|\vec{n}|^2 = 1 + 4 + 1 = 6$$

Compute $$P_0 \cdot \vec{n} + d$$: $$\frac{5}{3} - 2 \cdot \frac{5}{3} + \frac{8}{3} - 2 = \frac{5 - 10 + 8}{3} - 2 = 1 - 2 = -1$$

Find the image: $$P = \left(\frac{5}{3}, \frac{5}{3}, \frac{8}{3}\right) - \frac{2(-1)}{6}(1, -2, 1) = \left(\frac{5}{3}, \frac{5}{3}, \frac{8}{3}\right) + \frac{1}{3}(1, -2, 1)$$

$$P = \left(\frac{5+1}{3}, \frac{5-2}{3}, \frac{8+1}{3}\right) = (2, 1, 3)$$

Use the distance condition: $$Q = (6, -2, \alpha)$$, distance from $$P = (2, 1, 3)$$ is 13.

$$(6-2)^2 + (-2-1)^2 + (\alpha - 3)^2 = 169$$

$$16 + 9 + (\alpha - 3)^2 = 169$$

$$(\alpha - 3)^2 = 144$$

$$\alpha - 3 = \pm 12$$

$$\alpha = 15$$ or $$\alpha = -9$$

Since $$\alpha > 0$$, we get $$\alpha = 15$$.

The correct answer is 15.