Three, (25 points) Given a three-digit number $\overline{a b c}$ that satisfies $a+$ $b+c=a b c$. Find the sum of all such three-digit numbers.


Three, from the problem, we get that $a$, $b$, and $c$ are all non-zero, and without loss of generality, assume $a \geqslant b \geqslant c$.
When $b c>3$, $a+b+c=a b c>3 a$, which means $b+c>2 a$. This contradicts $b+c \leqslant 2 a$.

When $b c=3$, then $b=3, c=1$. In this case, $a=2$, which contradicts $a \geqslant b \geqslant c$.
Therefore, $b c<3$, which means $b=2, c=1$. In this case, $a=3$.
Thus, the three-digit numbers that satisfy the conditions are six: 123, 132, 213, 231, 312, 321, and their sum is 1332.
(Shen Xueming, Mudu High School, Jiangsu Province, 215101)