Simply

$\dfrac{1}{(7-x)(1+x)(1-x)} + \dfrac{3x^2-18x-22}{(x^2-1)(x-7)}+\dfrac{3}{x-2} = \dfrac{3}{(x-1)(x-2)}\\ \dfrac{1}{-1(x-7)(x+1)(-1)(x-1)} + \dfrac{3x^2-18x-22}{(x+1)(x-1)(x-7)} =\dfrac{3}{(x-1)(x-2)} - \dfrac{3}{x-2} \\ \dfrac{3x^2-18x-22+1}{(x+1)(x-1)(x-7)} =\dfrac{3 - 3(x-1)}{(x-1)(x-2)} \\ (3x^2 - 18x -21)(x-1)(x-2) = (3-3x+3)(x+1)(x-1)(x-7)\\ 3(x^2-6x-7)(x-1)(x-2) = (6-3x)(x+1)(x-1)(x-7)\\ 3(x-7)(x+1)(x-1)(x-2) = -3(x-2)(x+1)(x-1)(x-7)\\ 6(x-2)(x+1)(x-1)(x-7)=0\\ x=2,\;-1,\;1,\;7$

Notice that any of the roots given will result in one of the fractions in the equation to become undefined.

If $x=2,\;\dfrac{3}{x-2}$ is undefined.

If $x=1$ or $x=-1$ or $x=7,\;\dfrac{1}{(7-x)(1+x)(1-x)}$ is undefined.

Therefore, this equation has no real solutions. The answer is $\boxed{0}$