二、(25 points) Given positive integers $a, b, c$ satisfying the following conditions: $a>b>c,(a-b)(b$ $-c)(a-c c)=72$ and $a b c<100$. Find $a, b, c$.
II. Since $a-b, b-c, a-c$ are all positive integers, and $a-c=(a-b)+(b-c)$, when decomposing 72 into three factors, the only possibilities that satisfy the above conditions are $1,8,9$ or $8,1,9$. Therefore, we have
(1) $\left\{\begin{array}{l}a-b=1, \\ b-c=8, \\ a-c=9,\end{array}\right.$
(2) $\left\{\begin{array}{l}a-b=8, \\ b-c=1 \\ a-c=9 .\end{array}\right.$
From (1) and $a b c<100$, we get $a, b, c$ as $10,9,1$,
From (2) and $a b c<100$, we get $c=1$ or $c=2$. Thus, $a, b, c$ are $10,2,1$ or $11,3,2$.