Let $$P\left(\frac{2\sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}}\right)$$, Q, R and S be four points on the ellipse $$9x^2 + 4y^2 = 36$$. Let PQ and RS be mutually perpendicular and pass through the origin. If $$\frac{1}{(PQ)^2} + \frac{1}{(RS)^2} = \frac{p}{q}$$, where $$p$$ and $$q$$ are coprime, then $$p + q$$ is equal to

The ellipse $$9x^2 + 4y^2 = 36$$ can be written as $$\frac{x^2}{4} + \frac{y^2}{9} = 1$$ so that $$a^2 = 9,\ b^2 = 4$$ with semi-major axis $$a = 3$$ along the y-axis and semi-minor axis $$b = 2$$ along the x-axis. In this setting, PQ and RS denote two diameters passing through the origin that are perpendicular to each other.

First, one may parametrize the points on the ellipse by considering a line through the origin that makes an angle $$\theta$$ with the x-axis and has the equation $$y = x\tan\theta$$. Substituting this into the ellipse equation leads to $$\frac{x^2}{4} + \frac{x^2\tan^2\theta}{9} = 1$$.

Then if $$r$$ represents the distance from the origin to the intersection point, one has $$x = r\cos\theta$$ and $$y = r\sin\theta$$, which transforms the preceding relation into $$\frac{1}{r^2} = \frac{\cos^2\theta}{4} + \frac{\sin^2\theta}{9}$$.

Next, since PQ and RS are diameters, their lengths are given by $$PQ = 2r_1$$ and $$RS = 2r_2$$, where $$r_1$$ and $$r_2$$ are the corresponding radial distances for the directions $$\theta$$ and $$\theta + 90^\circ$$. Therefore, one finds

$$\frac{1}{(PQ)^2} + \frac{1}{(RS)^2} = \frac{1}{4r_1^2} + \frac{1}{4r_2^2} = \frac{1}{4}\Bigl(\frac{1}{r_1^2} + \frac{1}{r_2^2}\Bigr)\,.$$

Moreover, in the direction $$\theta$$ the relationship $$\frac{1}{r_1^2} = \frac{\cos^2\theta}{4} + \frac{\sin^2\theta}{9}$$ holds, while for the perpendicular direction $$\theta + 90^\circ$$ it becomes $$\frac{1}{r_2^2} = \frac{\sin^2\theta}{4} + \frac{\cos^2\theta}{9}$$. By adding these two expressions one obtains

$$\frac{1}{r_1^2} + \frac{1}{r_2^2} = \frac{\cos^2\theta + \sin^2\theta}{4} + \frac{\sin^2\theta + \cos^2\theta}{9} = \frac{1}{4} + \frac{1}{9} = \frac{13}{36}\,,$$

which is independent of $$\theta$$. Consequently,

$$\frac{1}{(PQ)^2} + \frac{1}{(RS)^2} = \frac{1}{4}\cdot\frac{13}{36} = \frac{13}{144}\,.$$

Since this result is of the form $$\frac{p}{q}$$ with $$p = 13$$ and $$q = 144$$ coprime, their sum is $$p + q = 13 + 144 = 157$$, giving the final answer as \boxed{157}.