$$
\begin{array}{l}
\text { Three, (25 points) Let three distinct prime numbers } a, b, c \text { satisfy } \\
a \text { divides }(3 b-c), b \text { divides }(a-c), c \text { divides }(2 a-7 b), \\
20<c<80 .
\end{array}
$$

Find all values of $a^{b} c$.
When $a>b, a>c$, we have:
$$
-7 c<-7 b<2 a-7 b<2 a<2 c \text {. }
$$

Let $2 a-7 b=k c$,
where, $k=-6,-5,-4,-3,-2,-1,0,1$.
Then $2 a=7 b+k c \Rightarrow 7 b=2 a-k c$,
$$
\begin{array}{l}
a|(3 b-c) \Leftrightarrow a|(3 \times 7 b-7 c) \\
\Leftrightarrow a \mid[3(2 a-k c)-7 c] \\
\Leftrightarrow a \mid[6 a-(3 k+7) c] \\
\Leftrightarrow a|(3 k+7) c \Leftrightarrow a|(3 k+7), \\
b|(a-c) \Leftrightarrow b|(2 a-2 c) \Leftrightarrow b \mid(7 b+k c-2 c) \\
\Leftrightarrow b|(k-2) c \Leftrightarrow b|(k-2) .
\end{array}
$$

When $k=-6$, from equations (2) and (3) we get
$$
a|(-11), b|(-8) \text {. }
$$

Thus, $a=11, b=2$.
From equation (1) we get $c=-\frac{4}{3}$ (discard).
When $k=-5$, from equations (2) and (3) we get
$a|(-8), b|(-7)$.
Thus, $a=2, b=7$.
From equation (1) we get $c=9$ (discard).
Similarly, when $k=-4,-3,-2,1$, there are no values that meet the conditions.

When $k=-1$, $a=2, b=3, c=17$, which does not satisfy $20<c<80$.
When $k=0$, $2 a=7 b$, then $217 b, 7 \mid 2 a$.
Thus, $b=2, a=7$.
From $a \mid(3 b-c)$, we get $3 b-c=a t, c=6-7 t$. Since $20<c<80$, we have $20<6-7 t<80,-10 \frac{4}{7}<t<-2$.
Only when $t=-5$, $c$ is a prime number, $c=41$. Therefore, $a^{b} c=7^{2} \times 41=2009$.
(Xie Wenxiao, Huangfeng Middle School, Hubei Province, 438000)