Three. (25 points) The roots of the equation $\left(m^{3}-2 m^{2}\right) x^{2}-$ $\left(m^{3}-3 m^{2}-4 m+8\right) x+12-4 m=0$ with respect to $x$ are all integers. Find the possible values of the real number $m$.

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When $m=0$, $-8 x+12=0$. $x$ is not an integer.
When $m=2$, $4 x+4=0 \Rightarrow x=-1$. Satisfied.
When $m \neq 0$ and $m \neq 2$, we have
\[
\begin{array}{l}
{\left[(m-2) x-(m-3) \right] m^{2} x+4=0 .} \\
x_{1}=\frac{m-3}{m-2}, \\
x_{2}=-\frac{4}{m^{2}} .
\end{array}
\]

From (1), we have $\left(x_{1}-1\right) m=2 x_{1}-3$.
If $x_{1}=1$, then $(m-2-m+3)\left(m^{2}+4\right)=0$, which is impossible.
So $m=\frac{2 x_{1}-3}{x_{1}-1}$.
Substituting into (2) gives $x_{2}=-1-\frac{4 x_{1}-5}{4 x_{1}^{2}-12 x_{1}+9}$.
Let $\frac{4 x_{1}-5}{4 x_{1}^{2}-12 x_{1}+9}=k \in \mathbf{Z}$. (Clearly $k \neq 0$, otherwise $x_{1}=\frac{5}{4} \notin \mathbf{Z}$) then
\[
\begin{array}{l}
4 k x_{1}^{2}-(12 k+4) x_{1}+9 k+5=0, \\
\Delta \geqslant 0 \Rightarrow k \geqslant-1 .
\end{array}
\]

When $k=-1$, $x_{1}=1$ (discard). Hence $k \geqslant 1$.
From $4 x_{1}-5 \geqslant 4 x_{1}^{2}-12 x_{1}+9$
\[
\Rightarrow 2 x_{1}^{2}-8 x_{1}+7 \leqslant 0 \text {. }
\]
Since $x_{1}$ is an integer, $\therefore x_{1}=2$.
Substituting into equation (3) gives $m=1$.
The other root $x_{2}=-4 \in \mathbf{Z}$ is satisfied.
In summary, the real number $m$ that satisfies the conditions is 1 or 2.