If $$\int_0^1 \frac{1}{\sqrt{3+x} + \sqrt{1+x}} \, dx = a + b\sqrt{2} + c\sqrt{3}$$, where $$a, b, c$$ are rational numbers, then $$2a + 3b - 4c$$ is equal to :

$$\int_0^1 \frac{1}{\sqrt{3+x} + \sqrt{1+x}} dx$$

Rationalize by multiplying by $$\frac{\sqrt{3+x} - \sqrt{1+x}}{\sqrt{3+x} - \sqrt{1+x}}$$:

$$= \int_0^1 \frac{\sqrt{3+x} - \sqrt{1+x}}{(3+x) - (1+x)} dx = \int_0^1 \frac{\sqrt{3+x} - \sqrt{1+x}}{2} dx$$

$$= \frac{1}{2}\int_0^1 (\sqrt{3+x} - \sqrt{1+x}) dx$$

$$= \frac{1}{2}\left[\frac{2}{3}(3+x)^{3/2} - \frac{2}{3}(1+x)^{3/2}\right]_0^1$$

$$= \frac{1}{3}\left[(4)^{3/2} - (2)^{3/2} - (3)^{3/2} + (1)^{3/2}\right]$$

$$= \frac{1}{3}\left[8 - 2\sqrt{2} - 3\sqrt{3} + 1\right]$$

$$= \frac{1}{3}(9 - 2\sqrt{2} - 3\sqrt{3})$$

$$= 3 - \frac{2\sqrt{2}}{3} - \sqrt{3}$$

So $$a = 3$$, $$b = -\frac{2}{3}$$, $$c = -1$$.

$$2a + 3b - 4c = 6 - 2 + 4 = 8$$

The answer is $$\boxed{8}$$, which corresponds to Option (4).