Three. (25 points) Given that $k$ is a constant, the quadratic equation in $x$
$$
\left(k^{2}-2 k\right) x^{2}+(4-6 k) x+8=0
$$

has solutions that are all integers. Find the value of $k$.
When $k=0$, the original equation becomes $4 x+8=0$, solving for $x$ gives $x=$ -2. Therefore, when $k=0$, the solutions to the original equation are all integers.

When $k=2$, the original equation becomes $-8 x+8=0$, solving for $x$ gives $x=$ 1. Therefore, when $k=2$, the solutions to the original equation are all integers.
When $k \neq 0$ or 2, the original equation becomes
$$
(k x-2)[(k-2) x-4]=0 \text {. }
$$

Solving for $x$ gives $x_{1}=\frac{2}{k}, x_{2}=\frac{4}{k-2}$.
From $x_{1}=\frac{2}{k}$, we get $k=\frac{2}{x_{1}}$.
Substituting $k=\frac{2}{x_{1}}$ into $x_{2}=\frac{4}{k-2}$, we get
$$
\begin{array}{l}
x_{1} x_{2}+2 x_{1}-x_{2}=0 . \\
\text { Therefore, }\left(x_{1}-1\right)\left(x_{2}+2\right)=-2=1 \times(-2)=2 \times(-1) .
\end{array}
$$

Since $x_{1}$ and $x_{2}$ are integers, $x_{1}-1$ and $x_{2}+2$ are also integers. Thus, we have
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ x _ { 1 } - 1 = 1 , } \\
{ x _ { 2 } + 2 = - 2 }
\end{array} \text { or } \left\{\begin{array}{l}
x_{1}-1=-2, \\
x_{2}+2=1
\end{array}\right.\right. \\
\text { or }\left\{\begin{array} { l } 
{ x _ { 1 } - 1 = 2 , } \\
{ x _ { 2 } + 2 = - 1 }
\end{array} \text { or } \left\{\begin{array}{l}
x_{1}-1=-1, \\
x_{2}+2=2 .
\end{array}\right.\right.
\end{array}
$$

Solving these, we get
$\left\{\begin{array}{l}x_{1}=2, \\ x_{2}=-4\end{array}\right.$ or $\left\{\begin{array}{l}x_{1}=-1, \\ x_{2}=-1\end{array}\right.$ or $\left\{\begin{array}{l}x_{1}=3, \\ x_{2}=-3\end{array}\right.$ or $\left\{\begin{array}{l}x_{1}=0, \\ x_{2}=0\end{array}\right.$ (discard). Therefore, $k=1,-2, \frac{2}{3}$.
In summary, the values of $k$ are $-2,0,1,2$ or $\frac{2}{3}$.
(Xie Wenxiao, Huanggang High School, Hubei Province, 438000)