Let $$P$$ be a point on the hyperbola $$H: \frac{x^2}{9} - \frac{y^2}{4} = 1$$, in the first quadrant such that the area of triangle formed by $$P$$ and the two foci of $$H$$ is $$2\sqrt{13}$$. Then, the square of the distance of $$P$$ from the origin is

Hyperbola: $$\frac{x^2}{9} - \frac{y^2}{4} = 1$$. So $$a = 3, b = 2, c = \sqrt{13}$$. Foci: $$(\pm\sqrt{13}, 0)$$.

Let $$P = (x_0, y_0)$$ in first quadrant on hyperbola.

Area of triangle with foci $$F_1(-\sqrt{13},0)$$ and $$F_2(\sqrt{13},0)$$:

Area = $$\frac{1}{2} \times 2\sqrt{13} \times y_0 = \sqrt{13} \cdot y_0 = 2\sqrt{13}$$.

So $$y_0 = 2$$. From hyperbola: $$\frac{x_0^2}{9} - \frac{4}{4} = 1 \Rightarrow \frac{x_0^2}{9} = 2 \Rightarrow x_0^2 = 18$$.

Distance from origin squared: $$x_0^2 + y_0^2 = 18 + 4 = 22$$.

The answer is Option (3): $$\boxed{22}$$.