If $$\frac{1}{\sqrt{1}+\sqrt{2}} + \frac{1}{\sqrt{2}+\sqrt{3}} + \ldots + \frac{1}{\sqrt{99}+\sqrt{100}} = m$$ and $$\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \ldots + \frac{1}{99 \cdot 100} = n$$, then the point $$(m, n)$$ lies on the line

We first compute $$m = \sum_{k=1}^{99} \frac{1}{\sqrt{k} + \sqrt{k+1}} = \sum_{k=1}^{99} \frac{\sqrt{k+1} - \sqrt{k}}{(\sqrt{k+1})^2 - (\sqrt{k})^2} = \sum_{k=1}^{99} (\sqrt{k+1} - \sqrt{k}).$$ This telescopes to $$m = \sqrt{100} - \sqrt{1} = 10 - 1 = 9.$$

Next, we evaluate $$n = \sum_{k=1}^{99} \frac{1}{k(k+1)} = \sum_{k=1}^{99} \left(\frac{1}{k} - \frac{1}{k+1}\right) = 1 - \frac{1}{100} = \frac{99}{100}.$$

Substituting $$(m,n) = (9, 99/100)$$ into the given options shows that Option (2) satisfies

$$11x - 100y = 11(9) - 100\bigl(99/100\bigr) = 99 - 99 = 0.$$

Therefore, the correct answer is Option (2): $$11x - 100y = 0.$$