Three. (25 points) Given the equation
$$
\left(m^{2}-1\right) x^{2}-3(3 m-1) x+18=0
$$

has two positive integer roots, where $m$ is an integer.
(1) Find the value of $m$;
(2) The sides of $\triangle A B C$, $a$, $b$, and $c$, satisfy $c=2 \sqrt{3}$, $m^{2}+a^{2} m-8 a=0$, $m^{2}+b^{2} m-8 b=0$, find the area of $\triangle A B C$.
Three, (1) It is easy to know that the two roots of the equation are
$$
x_{1}=\frac{6}{m+1}, x_{2}=\frac{3}{m-1} \text {. }
$$

From the problem, we have
$$
\left\{\begin{array}{l}
m+1=1,2,3,6, \\
m-1=1,3 .
\end{array}\right.
$$

Thus, $m=2$.
(2) According to the problem, $a$ and $b$ are both roots of the equation
$$
x^{2}-4 x+2=0
$$

(i) $a \neq b, c=2 \sqrt{3}$.
By $a^{2}+b^{2}=(a+b)^{2}-2 a b=16-4=12=c^{2}$, we know $\angle C=90^{\circ}$, and $S_{\triangle A B C}=\frac{1}{2} a b=1$.
(ii) $a=b=2-\sqrt{2}, c=2 \sqrt{3}$.
By $2(2-\sqrt{2})2 \sqrt{3}$, we know a triangle can be formed, and
$S_{\triangle A B C}=\frac{1}{2} \times 2 \sqrt{3} \times \sqrt{(2+\sqrt{2})^{2}-(\sqrt{3})^{2}}$ $=\sqrt{9+12 \sqrt{2}}$.
In summary, the area of $\triangle A B C$ is 1 or $\sqrt{9+12 \sqrt{2}}$.