Three. (25 points) Find all positive integers $a$ such that the quadratic equation
$$
a x^{2}+2(2 a-1) x+4(a-3)=0
$$

has at least one integer root.
Three, the original equation can be transformed into $(x+2)^{2} a=2 x+12$.
It is easy to see that $x \neq-2$, at this time $a=\frac{2 x+12}{(x+2)^{2}}$.
Since $a$ is a positive integer, i.e., $\frac{2 x+12}{(x+2)^{2}} \geqslant 1$ is a positive integer.
Also, $(x+2)^{2}>0$, then
$$
(x+2)^{2} \leqslant 2 x+12 \text {, }
$$

i.e., $x^{2}+2 x-8 \leqslant 0$.
Solving this, we get $-4 \leqslant x \leqslant 2$.
Since $x \neq-2$, and $x$ is an integer, $x$ can only take -4, $-3,-1,0,1,2$. Substituting these values into the expression for $a$, we get
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ x = - 4 , } \\
{ a = 1 ; }
\end{array} \quad \left\{\begin{array} { l } 
{ x = - 3 , } \\
{ a = 6 ; }
\end{array} \quad \left\{\begin{array}{l}
x=-1, \\
a=10 ;
\end{array}\right.\right.\right. \\
\left\{\begin{array} { l } 
{ x = 0 , } \\
{ a = 3 ; }
\end{array} \quad \left\{\begin{array} { l } 
{ x = 1 , } \\
{ a = \frac { 1 4 } { 9 } ; }
\end{array} \quad \left\{\begin{array}{l}
x=2, \\
a=1 .
\end{array}\right.\right.\right. \\
\end{array}
$$

Thus, there are 4 positive integer values of $a$ that satisfy the requirements, $a=1$, $6,10,3$.