Three. (25 points) Determine all rational numbers $r$ such that the equation $r x^{2}+(r+2) x+3 r-2=0$ has roots and only integer roots.

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Three. (25 points) Determine all rational numbers $r$ such that the equation $r x^{2}+(r+2) x+3 r-2=0$ has roots and only integer roots.
If $r=0$, then the equation becomes $2 x-2=0$, solving for $x$ gives $x=1$. If $r \neq 0$, let the two roots of the equation be $x_{1} 、 x_{2}\left(x_{1} \leqslant x_{2}\right)$, then $x_{1}+x_{2}=-\frac{r+2}{r}, x_{1} x_{2}=\frac{3 r-2}{r}$.
We have $\quad x_{1} x_{2}-\left(x_{1}+x_{2}\right)=\frac{3 r-2}{r}+\frac{r+2}{r}=4$.
Then $\left(x_{1}-1\right)\left(x_{2}-1\right)=5$.
Also, since $x_{1}<x_{2}$ and $x_{1} 、 x_{2}$ are integers,
Solving gives $\left\{\begin{array}{l}x_{1}=2 \\ x_{2}=6\end{array}\right.$ or $\left\{\begin{array}{l}x_{1}=-4, \\ x_{2}=0 .\end{array}\right.$
Thus, $r=-\frac{2}{9}$ or $r=\frac{2}{3}$.
Therefore, all the rational numbers $r$ sought are $0,-\frac{2}{9}, \frac{2}{3}$.