Упростите выражение: a) $$\sqrt{6}(\sqrt{12} - \sqrt{6}) - \frac{3}{2}\sqrt{32}$$; б) $$(\sqrt{7}-\sqrt{3})^2$$.

a) $$\sqrt{6}(\sqrt{12} - \sqrt{6}) - \frac{3}{2}\sqrt{32} = \sqrt{6}(\sqrt{4\times3} - \sqrt{6}) - \frac{3}{2}\sqrt{16\times2} = \sqrt{6}(2\sqrt{3} - \sqrt{6}) - \frac{3}{2}4\sqrt{2} = 2\sqrt{18} - 6 - 6\sqrt{2} = 2\sqrt{9\times2} - 6 - 6\sqrt{2} = 2\times3\sqrt{2} - 6 - 6\sqrt{2} = 6\sqrt{2} - 6 - 6\sqrt{2} = -6$$ \textbf{Ответ: -6} б) $$(\sqrt{7}-\sqrt{3})^2 = (\sqrt{7})^2 - 2\sqrt{7}\sqrt{3} + (\sqrt{3})^2 = 7 - 2\sqrt{21} + 3 = 10 - 2\sqrt{21}$$ \textbf{Ответ: $$10 - 2\sqrt{21}$$}