a) 6\sqrt{3} + \sqrt{27} - 3\sqrt{75} = 6\sqrt{3} + \sqrt{9 \cdot 3} - 3\sqrt{25 \cdot 3} = 6\sqrt{3} + 3\sqrt{3} - 3 \cdot 5\sqrt{3} = 6\sqrt{3} + 3\sqrt{3} - 15\sqrt{3} = (6+3-15)\sqrt{3} = -6\sqrt{3}.
б) (\sqrt{50} - 2\sqrt{2})\sqrt{2} = (\sqrt{25 \cdot 2} - 2\sqrt{2})\sqrt{2} = (5\sqrt{2} - 2\sqrt{2})\sqrt{2} = 3\sqrt{2} \cdot \sqrt{2} = 3 \cdot 2 = 6.
в) (2-\sqrt{3})^2 = 2^2 - 2 \cdot 2 \cdot \sqrt{3} + (\sqrt{3})^2 = 4 - 4\sqrt{3} + 3 = 7 - 4\sqrt{3}.
Ответ: а) -6√3; б) 6; в) 7 - 4√3