a) $$3\sqrt{2} = \sqrt{3^2 \cdot 2} = \sqrt{9 \cdot 2} = \sqrt{18}$$
б) $$a\sqrt{3} = -\sqrt{a^2 \cdot 3} = -\sqrt{3a^2}$$, т.к. $$a < 0$$
в) $$-x\sqrt{\frac{2}{x}} = -\sqrt{x^2 \cdot \frac{2}{x}} = -\sqrt{2x}$$
Ответ:
a) $$\boxed{\sqrt{18}}$$
б) $$\boxed{-\sqrt{3a^2}}$$
в) $$\boxed{-\sqrt{2x}}$$