Let the latus rectum of the hyperbola $$\frac{x^2}{9} - \frac{y^2}{b^2} = 1$$ subtend an angle of $$\frac{\pi}{3}$$ at the centre of the hyperbola. If $$b^2$$ is equal to $$\frac{l}{m}(1 + \sqrt{n})$$, where $$l$$ and $$m$$ are co-prime numbers, then $$l^2 + m^2 + n^2$$ is equal to __________.

For the hyperbola $$\frac{x^2}{9} - \frac{y^2}{b^2} = 1$$, we have $$a = 3$$.

The semi-latus rectum is $$l = \frac{b^2}{a} = \frac{b^2}{3}$$. The end of the latus rectum is at $$(ae, \frac{b^2}{a})$$, where $$c = ae = \sqrt{9 + b^2}$$.

The latus rectum subtends angle $$\frac{\pi}{3}$$ at the centre. The half-angle at centre is $$\frac{\pi}{6}$$.

$$\tan\frac{\pi}{6} = \frac{b^2/3}{\sqrt{9+b^2}}$$

$$\frac{1}{\sqrt{3}} = \frac{b^2}{3\sqrt{9+b^2}}$$

$$3\sqrt{9+b^2} = \sqrt{3} b^2$$

Squaring: $$9(9+b^2) = 3b^4$$

$$81 + 9b^2 = 3b^4$$

$$3b^4 - 9b^2 - 81 = 0$$

$$b^4 - 3b^2 - 27 = 0$$

Using the quadratic formula with $$u = b^2$$:

$$u = \frac{3 + \sqrt{9 + 108}}{2} = \frac{3 + \sqrt{117}}{2} = \frac{3 + 3\sqrt{13}}{2}$$

So $$b^2 = \frac{3}{2}(1 + \sqrt{13})$$.

Here $$l = 3, m = 2, n = 13$$ (with $$l, m$$ coprime).

$$l^2 + m^2 + n^2 = 9 + 4 + 169 = 182$$.

Therefore, the answer is $$\boxed{182}$$.