jmc

We can move the $\frac{3}{x-2}$ to the right-hand side. We can also combine the first two fractions, to get $$\begin{array}{rl}\frac{3x^2-18x-21}{(x-1)(x+1)(x-7)}& =\frac{3}{(x-1)(x-2)}-\frac{3}{x-2}\\ & =\frac{3-3(x-1)}{(x-1)(x-2)}\\ & =\frac{6-3x}{(x-1)(x-2)}\\ & =\frac{3(2-x)}{(x-1)(x-2)}\\ & =-\frac{3}{x-1}.\end{array}$$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 $\boxed{0}.$