Let the lines $$y + 2x = \sqrt{11} + 7\sqrt{7}$$ and $$2y + x = 2\sqrt{11} + 6\sqrt{7}$$ be normal to a circle $$C : (x-h)^2 + (y-k)^2 = r^2$$. If the line $$\sqrt{11}y - 3x = \frac{5\sqrt{77}}{3} + 11$$ is tangent to the circle $$C$$, then the value of $$(5h - 8k)^2 + 5r^2$$ is equal to ______

The center of the circle $$(h, k)$$ lies on both normals:

$$k + 2h = \sqrt{11} + 7\sqrt{7} \quad \cdots (1)$$

$$2k + h = 2\sqrt{11} + 6\sqrt{7} \quad \cdots (2)$$

From (1): $$k = \sqrt{11} + 7\sqrt{7} - 2h$$. Substituting into (2):

$$2(\sqrt{11} + 7\sqrt{7} - 2h) + h = 2\sqrt{11} + 6\sqrt{7}$$

$$2\sqrt{11} + 14\sqrt{7} - 4h + h = 2\sqrt{11} + 6\sqrt{7}$$

$$-3h = -8\sqrt{7}$$

$$h = \frac{8\sqrt{7}}{3}$$

$$k = \sqrt{11} + 7\sqrt{7} - \frac{16\sqrt{7}}{3} = \sqrt{11} + \frac{21\sqrt{7} - 16\sqrt{7}}{3} = \sqrt{11} + \frac{5\sqrt{7}}{3}$$

The tangent line is $$\sqrt{11}y - 3x = \frac{5\sqrt{77}}{3} + 11$$, or $$3x - \sqrt{11}y + \frac{5\sqrt{77}}{3} + 11 = 0$$.

Multiply through by 3: $$9x - 3\sqrt{11}y + 5\sqrt{77} + 33 = 0$$.

The distance from center $$(h, k)$$ to this line equals $$r$$:

$$r = \frac{|9h - 3\sqrt{11}k + 5\sqrt{77} + 33|}{\sqrt{81 + 99}} = \frac{|9h - 3\sqrt{11}k + 5\sqrt{77} + 33|}{\sqrt{180}} = \frac{|9h - 3\sqrt{11}k + 5\sqrt{77} + 33|}{6\sqrt{5}}$$

Computing $$9h - 3\sqrt{11}k$$:

$$9h = 9 \cdot \frac{8\sqrt{7}}{3} = 24\sqrt{7}$$

$$3\sqrt{11}k = 3\sqrt{11}\left(\sqrt{11} + \frac{5\sqrt{7}}{3}\right) = 3 \cdot 11 + 5\sqrt{77} = 33 + 5\sqrt{77}$$

$$9h - 3\sqrt{11}k + 5\sqrt{77} + 33 = 24\sqrt{7} - 33 - 5\sqrt{77} + 5\sqrt{77} + 33 = 24\sqrt{7}$$

$$r = \frac{24\sqrt{7}}{6\sqrt{5}} = \frac{4\sqrt{7}}{\sqrt{5}}$$

$$r^2 = \frac{112}{5}$$

Now computing $$(5h - 8k)^2 + 5r^2$$:

$$5h = \frac{40\sqrt{7}}{3}, \quad 8k = 8\sqrt{11} + \frac{40\sqrt{7}}{3}$$

$$5h - 8k = \frac{40\sqrt{7}}{3} - 8\sqrt{11} - \frac{40\sqrt{7}}{3} = -8\sqrt{11}$$

$$(5h - 8k)^2 = 64 \times 11 = 704$$

$$5r^2 = 5 \times \frac{112}{5} = 112$$

$$(5h - 8k)^2 + 5r^2 = 704 + 112 = 816$$

Hence the answer is $$\boxed{816}$$.