Three, (25 points) Find all integer triples $(a, b, c)$
such that
$$
a^{3}+b^{3}+c^{3}-3 a b c=2011(a \geqslant b \geqslant c) .
$$
$$
\begin{array}{l}
(a+b+c)\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right] \\
=2 \times 2011 .
\end{array}
$$

Since $(a-b)^{2}+(b-c)^{2}+(c-a)^{2}$ is always even, we have:
$$
\left\{\begin{array}{l}
a+b+c=1, \\
(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=4022 ;
\end{array}\right.
$$

or $\left\{\begin{array}{l}a+b+c=2011, \\ (a-b)^{2}+(b-c)^{2}+(c-a)^{2}=2 .\end{array}\right.$
For the first system of equations, let $x=a-b, y=b-c$.
Then $x^{2}+y^{2}+(x+y)^{2}=4022$
$\Rightarrow x^{2}+y^{2}+x y=2011$
$\Rightarrow(2 x+y)^{2}=8044-3 y^{2}$.
Also, $y^{2} \leqslant 2011 \Rightarrow 0 \leqslant y \leqslant 44$.
Upon inspection, only $y=10,39$ make $8044-3 y^{2}$ a perfect square.
Thus, $(x, y)=(10,39),(39,10)$.
Therefore, $(a, b, c)=(20,10,-29)$.
For the second system of equations, $a-b, b-c, a-c$ include two 1s and one 0.
When $a=b=c+1$,
$$
2(c+1)+c=2011
$$

has no solution;
When $a-1=b=c$,
$$
a+2(a-1)=2011 \Rightarrow a=671 \text {. }
$$

Thus, $b=670, c=670$.
Therefore, the integer solutions for the triple are
$(20,10,-29),(671,670,670)$.
(Wu Jinping, Hengche Middle School, Wuchang County, Hubei, 435300; Weng Hongtao, Chen Hong, Jinhe Middle School, Jinchun County, Hubei, 435300)