The value of $$\frac{3+\sqrt{6}}{8\sqrt{3}-2\sqrt{12}-\sqrt{32}+\sqrt{50}-\sqrt{27}}$$ is

Simplifying the denominator, $$-2\sqrt{12}=-4\sqrt{3}$$, $$-\sqrt{32}+\sqrt{50}=-4\sqrt{2}+5\sqrt{2}=\sqrt{2}$$ and $$-\sqrt{27}=-3\sqrt{3}$$, so the denominator becomes $$8\sqrt{3}-4\sqrt{3}-3\sqrt{3}+\sqrt{2}=\sqrt{3}+\sqrt{2}$$. Rationalizing, $$\frac{3+\sqrt{6}}{\sqrt{3}+\sqrt{2}}\times\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\frac{3\sqrt{3}-3\sqrt{2}+\sqrt{18}-\sqrt{12}}{1}=\frac{3\sqrt{3}-3\sqrt{2}+3\sqrt{2}-2\sqrt{3}}{1}=\sqrt{3}$$.