Let $$S = \{x : x \in \mathbb{R}$$ and $$\sqrt{3} + \sqrt{2}^{x^2-4} + \sqrt{3} - \sqrt{2}^{x^2-4} = 10\}$$. Then $$nS$$ is equal to

Let $$t = (\sqrt{3} + \sqrt{2})^{x^2 - 4}$$.

Since $$(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = 3 - 2 = 1$$, we have $$(\sqrt{3} - \sqrt{2}) = \frac{1}{\sqrt{3} + \sqrt{2}}$$.

Therefore: $$(\sqrt{3} - \sqrt{2})^{x^2-4} = \frac{1}{t}$$

The equation becomes:

Note that $$(\sqrt{3} + \sqrt{2})^2 = 3 + 2\sqrt{6} + 2 = 5 + 2\sqrt{6}$$.

Case 1: $$t = 5 + 2\sqrt{6} = (\sqrt{3} + \sqrt{2})^2$$

$$x^2 - 4 = 2 \Rightarrow x^2 = 6 \Rightarrow x = \pm\sqrt{6}$$

Case 2: $$t = 5 - 2\sqrt{6} = (\sqrt{3} + \sqrt{2})^{-2}$$

$$x^2 - 4 = -2 \Rightarrow x^2 = 2 \Rightarrow x = \pm\sqrt{2}$$

Total solutions: $$\{-\sqrt{6}, -\sqrt{2}, \sqrt{2}, \sqrt{6}\}$$

Therefore, $$n(S) = 4$$.