Let $$\alpha, \beta; \alpha > \beta$$, be the roots of the equation $$x^2 - \sqrt{2}x - \sqrt{3} = 0$$. Let $$P_n = \alpha^n - \beta^n, n \in \mathbb{N}$$. Then $$(11\sqrt{3} - 10\sqrt{2})P_{10} + (11\sqrt{2} + 10)P_{11} - 11P_{12}$$ is equal to

We need to simplify the expression $$(11\sqrt{3}-10\sqrt{2})P_{10} + (11\sqrt{2}+10)P_{11} - 11P_{12}.$$

Since $$\alpha, \beta$$ are roots of $$x^2 - \sqrt{2}x - \sqrt{3} = 0$$, we have $$\alpha + \beta = \sqrt{2}$$ and $$\alpha\beta = -\sqrt{3}$$. The sequence $$P_n = \alpha^n - \beta^n$$ therefore satisfies the recurrence $$P_n = \sqrt{2}P_{n-1} + \sqrt{3}P_{n-2}$$, because $$\alpha^n = \sqrt{2}\alpha^{n-1} + \sqrt{3}\alpha^{n-2}$$ (and similarly for $$\beta$$).

In particular, $$P_{12} = \sqrt{2}P_{11} + \sqrt{3}P_{10}$$, so $$11P_{12} = 11\sqrt{2}P_{11} + 11\sqrt{3}P_{10}$$.

Substituting this into the original expression gives

$$(11\sqrt{3}-10\sqrt{2})P_{10} + (11\sqrt{2}+10)P_{11} - \bigl(11\sqrt{2}P_{11} + 11\sqrt{3}P_{10}\bigr).$$

Collecting like terms, the coefficient of $$P_{10}$$ is $$(11\sqrt{3}-10\sqrt{2}) - 11\sqrt{3} = -10\sqrt{2}$$ and the coefficient of $$P_{11}$$ is $$(11\sqrt{2}+10) - 11\sqrt{2} = 10$$. Thus the expression simplifies to $$-10\sqrt{2}P_{10} + 10P_{11}$$.

Factoring out 10 gives $$10(P_{11} - \sqrt{2}P_{10})$$. Using the recurrence backwards, $$P_{11} = \sqrt{2}P_{10} + \sqrt{3}P_9$$, so $$P_{11} - \sqrt{2}P_{10} = \sqrt{3}P_9$$ and hence the result is $$10\sqrt{3}P_9$$.

The correct answer is Option 1: $$10\sqrt{3}P_9$$.