Let S be the focus of the hyperbola $$\frac{x^2}{3} - \frac{y^2}{5} = 1$$, on the positive x-axis. Let C be the circle with its centre at $$A(\sqrt{6}, \sqrt{5})$$ and passing through the point S. If O is the origin and SAB is a diameter of C, then the square of the area of the triangle OSB is equal to _____

For the hyperbola $$\frac{x^2}{3} - \frac{y^2}{5} = 1$$: $$a^2 = 3, b^2 = 5$$, so $$c^2 = a^2 + b^2 = 8$$, $$c = 2\sqrt{2}$$.

Focus on positive x-axis: $$S = (2\sqrt{2}, 0)$$.

Circle C has centre $$A(\sqrt{6}, \sqrt{5})$$ and passes through S. Radius = $$|AS|$$.

$$|AS|^2 = (2\sqrt{2}-\sqrt{6})^2 + (0-\sqrt{5})^2 = (8 - 4\sqrt{12} + 6) + 5 = 19 - 4\sqrt{12}$$

$$= 19 - 8\sqrt{3}$$

SAB is a diameter, so B is the diametrically opposite point of S through centre A.

$$B = 2A - S = (2\sqrt{6} - 2\sqrt{2}, 2\sqrt{5})$$.

Now find the area of triangle OSB, where $$O = (0,0)$$, $$S = (2\sqrt{2}, 0)$$, $$B = (2\sqrt{6}-2\sqrt{2}, 2\sqrt{5})$$.

Area = $$\frac{1}{2}|x_S \cdot y_B - x_B \cdot y_S|$$

$$= \frac{1}{2}|2\sqrt{2} \cdot 2\sqrt{5} - (2\sqrt{6}-2\sqrt{2}) \cdot 0|$$

$$= \frac{1}{2} \cdot 4\sqrt{10} = 2\sqrt{10}$$

Square of area = $$(2\sqrt{10})^2 = 40$$.

The answer is $$\boxed{40}$$.