12. Given the three sides of $\triangle A B C$ are $a, b, c$, and they satisfy $a b c=2(a-1)(b-1)(c-1)$, does there exist a $\triangle A B C$ with all sides being integers? If it exists, find the lengths of the three sides; if not, explain the reason.
Solution: Without loss of generality, let $a \geq b \geq c$, it is obvious that $c \geq 2$.
5 points
If $c \geq 5$, then we have
$$
\frac{1}{a} \leq \frac{1}{b} \leq \frac{1}{c} \leq \frac{1}{5}
$$

From $a b c=2(a-1)(b-1)(c-1)$, we get
$$
\frac{1}{2}=\left(1-\frac{1}{a}\right)\left(1-\frac{1}{b}\right)\left(1-\frac{1}{c}\right) \geq\left(\frac{4}{5}\right)^{3}
$$

This is a contradiction. Therefore, $c$ can only be $2,3,4$.
10 points
(1) If $c=2$, then $a b=(a-1)(b-1)$, which gives $a+b=1$. Since $a \geq b \geq 2$, there is no solution;
(2) If $c=3$, then $3 a b=4(a-1)(b-1)$, which simplifies to $(a-4)(b-4)=12$. Given $a \geq b \geq 3$, we have
$$
\left\{\begin{array}{l}
a - 4 = 12 \\
b - 4 = 1
\end{array} \text {, or } \left\{\begin{array}{l}
a - 4 = 6 \\
b - 4 = 2
\end{array} \text {, or } \left\{\begin{array}{l}
a - 4 = 4 \\
b - 4 = 3
\end{array} \text {, solving to } \left\{\begin{array}{l}
a = 16 \\
b = 5
\end{array} \text {, or } \left\{\begin{array}{l}
a = 10 \\
b = 6
\end{array} \text {, or } \left\{\begin{array}{l}
a = 8 \\
b = 7
\end{array}\right.\right.\right.\right.\right.\right.
$$

Among these, the only solution that can form a triangle is $a=8, b=7, c=3$;
15 points
(3) If $c=4$, then $2 a b=3(a-1)(b-1)$, which simplifies to $(a-3)(b-3)=6$. Given $a \geq b \geq 4$, we have
$$
\left\{\begin{array}{l}
a - 3 = 6 \\
b - 3 = 1
\end{array} \text {, or } \left\{\begin{array}{l}
a - 3 = 3 \\
b - 3 = 2
\end{array} \text {, solving to } \left\{\begin{array}{l}
a = 9 \\
b = 4
\end{array} \text {, or } \left\{\begin{array}{l}
a = 6 \\
b = 5
\end{array}\right.\right.\right.\right.
$$

Among these, the only solution that can form a triangle is $a=6, b=5, c=4$.
In summary, there exist triangles $\triangle ABC$ with integer side lengths that satisfy the conditions, with side lengths of $3,7,8$ or $4,5,6$.
20 points