If the circles $$x^2 + y^2 + 6x + 8y + 16 = 0$$ and $$x^2 + y^2 + 2(3 - \sqrt{3})x + 2(4 - \sqrt{6})y = k + 6\sqrt{3} + 8\sqrt{6}$$, $$k > 0$$, touch internally at the point $$P(\alpha, \beta)$$, then $$(\alpha + \sqrt{3})^2 + (\beta + \sqrt{6})^2$$ is equal to ______.

Circle 1: $$x^2 + y^2 + 6x + 8y + 16 = 0$$ has center $$C_1 = (-3,-4)$$ and radius $$r_1 = \sqrt{9 + 16 - 16} = 3$$.

Circle 2: $$x^2 + y^2 + 2(3-\sqrt{3})x + 2(4-\sqrt{6})y = k + 6\sqrt{3} + 8\sqrt{6}$$ has center $$C_2 = \bigl(-(3-\sqrt{3}),\,-(4-\sqrt{6})\bigr) = (-3+\sqrt{3}, -4+\sqrt{6})$$.

For the radius $$r_2$$ of the second circle we have

$$r_2^2 = (3-\sqrt{3})^2 + (4-\sqrt{6})^2 + k + 6\sqrt{3} + 8\sqrt{6}$$ $$= (12 - 6\sqrt{3}) + (22 - 8\sqrt{6}) + k + 6\sqrt{3} + 8\sqrt{6} = 34 + k$$

The distance between the centers is

$$|C_1C_2| = \sqrt{(\sqrt{3})^2 + (\sqrt{6})^2} = \sqrt{3 + 6} = 3$$

For internal tangency we set

$$|\sqrt{34+k} - 3| = 3$$

Since $$k > 0$$ we have $$\sqrt{34+k} = 6\implies k = 2$$, so $$r_2 = 6$$.

The point of tangency $$P(\alpha,\beta)$$ lies on the line $$C_1C_2$$ at distance $$r_1 = 3$$ from $$C_1$$ in the direction opposite to $$C_2$$, hence

$$P = C_1 - r_1 \cdot \frac{\vec{C_2 - C_1}}{|C_1C_2|} = (-3, -4) - 3 \cdot \frac{(\sqrt{3}, \sqrt{6})}{3} = (-3 - \sqrt{3}, -4 - \sqrt{6})$$

Finally,

$$(\alpha + \sqrt{3})^2 + (\beta + \sqrt{6})^2 = (-3-\sqrt{3}+\sqrt{3})^2 + (-4-\sqrt{6}+\sqrt{6})^2 = (-3)^2 + (-4)^2 = 9 + 16 = 25$$

The answer is $$\boxed{25}$$.