The sum of all rational terms in the expansion of $$(1+2^{1/2}+3^{1/2})^{6}$$ is equal to

We need to find the sum of all rational terms in the expansion of $$(1 + \sqrt{2} + \sqrt{3})^6$$.

By the multinomial theorem:

$$(1 + \sqrt{2} + \sqrt{3})^6 = \sum_{a+b+c=6} \frac{6!}{a!\,b!\,c!} \cdot 1^a \cdot (\sqrt{2})^b \cdot (\sqrt{3})^c$$

$$= \sum_{a+b+c=6} \frac{6!}{a!\,b!\,c!} \cdot 2^{b/2} \cdot 3^{c/2}$$

A term is rational when $$2^{b/2} \cdot 3^{c/2}$$ is rational, which requires both $$b$$ and $$c$$ to be even.

Let $$b = 2p$$ and $$c = 2q$$ where $$p, q \geq 0$$ and $$a + 2p + 2q = 6$$, so $$a = 6 - 2p - 2q$$.

We need $$a \geq 0$$, so $$p + q \leq 3$$:

$$S = 1 + 30 + 45 + 60 + 540 + 135 + 8 + 180 + 270 + 27 = 1296$$

The answer is $$\boxed{1296}$$.