We can move the $ \frac{3}{x - 2} $ to the right-hand side. We can also combine the first two fractions, to get \begin{align*} \frac{3x^2 - 18x - 21}{(x - 1)(x + 1)(x - 7)} &= \frac{3}{(x - 1)(x - 2)} - \frac{3}{x - 2} \\ &= \frac{3 - 3(x - 1)}{(x - 1)(x - 2)} \\ &= \frac{6 - 3x}{(x - 1)(x - 2)} \\ &= \frac{3(2 - x)}{(x - 1)(x - 2)} \\ &= -\frac{3}{x - 1}.\end{align*}Note that $ 3x^2 - 18x - 21 $ factors as $ 3(x + 1)(x - 7), $ so \[\frac{3(x + 1)(x - 7)}{(x - 1)(x + 1)(x - 7)} = -\frac{3}{x - 1}.\]This simplifies to \[\frac{3}{x - 1} = -\frac{3}{x - 1}.\]This can never occur, so the number of solutions is $ \boxed{0} $.