Three, if the equation with respect to $x$
$$
a x^{2}+2(a-3) x+(a-2)=0
$$

has at least one integer solution, and $a$ is an integer, find $a$.


When $a=0$, the known equation becomes $-6 x-2=0$, which has no integer solutions.

When $a \neq 0$, the known equation has at least one integer solution, which requires the discriminant $\Delta=4(a-3)^{2}-4 a(a-2)=4(9-4 a)$ to be a perfect square, thus $9-4 a$ must be a perfect square.
Let $9-4 a=s^{2}(s$ be a positive odd number, and $s \neq 3)$,
then $a=\frac{9-s^{2}}{4}$.
$$
\begin{array}{l}
x_{1,2}=\frac{-2(a-3) \pm 2 s}{2 a}=-1+\frac{4(3 \pm s)}{9-s^{2}}, \\
x_{1}=-1+\frac{4}{3+s}, x_{2}=-1+\frac{4}{3-s} .
\end{array}
$$

To make $x_{1}$ an integer, if $s$ is a positive odd number, $s$ can only be $1, a=2$.
To make $x_{2}$ an integer, i.e., $(3-s) \mid 4$, $s$ can only be $1, 5, 7$.
When $s=5$ or 7, $a=-4$ or -10.
In summary, the values of $a$ are $2, -4, -10$.