smc

Using Simon's Favorite Factoring Trick, we can rewrite the three equations as follows: $$\begin{array}{rl}(a+1)(b+1)& =525\\ (b+1)(c+1)& =147\\ (c+1)(d+1)& =105\end{array}$$ Let $(e,f,g,h)=(a+1,b+1,c+1,d+1)$. We get: $$\begin{array}{rl}ef& =3\cdot 5\cdot 5\cdot 7\\ fg& =3\cdot 7\cdot 7\\ gh& =3\cdot 5\cdot 7\end{array}$$ Clearly $7^2$ divides $fg$. On the other hand, $7^2$ can not divide $f$, as it then would divide $ef$. Similarly, $7^2$ can not divide $g$. Hence $7$ divides both $f$ and $g$. This leaves us with only two cases: $(f,g)=(7,21)$ and $(f,g)=(21,7)$. The first case solves to $(e,f,g,h)=(75,7,21,5)$, which gives us $(a,b,c,d)=(74,6,20,4)$, but then $abcd\ne 8!$. We do not need to multiply, it is enough to note e.g. that the left hand side is not divisible by $7$. (Also, a - d equals $70$ in this case, which is way too large to fit the answer choices.) The second case solves to $(e,f,g,h)=(25,21,7,15)$, which gives us a valid quadruple $(a,b,c,d)=(24,20,6,14)$, and we have $a-d=24-14=\boxed{10}$.