Five. (20 points) Given that the side lengths 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 side lengths being integers? If so, find the three side lengths; if not, explain the reason.
Let's assume $a \geqslant b \geqslant c$. Clearly, $c \geqslant 2$.
If $c \geqslant 5$, then $\frac{1}{a} \leqslant \frac{1}{b} \leqslant \frac{1}{c} \leqslant \frac{1}{5}$.
From $a b c=2(a-1)(b-1)(c-1)$, we get
$\frac{1}{2}=\left(1-\frac{1}{a}\right)\left(1-\frac{1}{b}\right)\left(1-\frac{1}{c}\right) \geqslant\left(\frac{4}{5}\right)^{3}$,
which is a contradiction.
Therefore, $c$ can only be $2, 3, 4$.
(1) If $c=2$.
Then $a b=(a-1)(b-1) \Rightarrow a+b=1$.
Since $a \geqslant b \geqslant 2$, there is no solution.
(2) If $c=3$.
Then $3 a b=4(a-1)(b-1)$
$\Rightarrow(a-4)(b-4)=12$.
Since $a \geqslant b \geqslant 3$, the solutions are
$(a, b)=(16,5),(10,6),(8,7)$.
Upon verification, the only solution that forms a triangle is
$$
a=8, b=7, c=3 \text {. }
$$
(3) If $c=4$.

Then $2 a b=3(a-1)(b-1)$
$\Rightarrow(a-3)(b-3)=6$.
Since $a \geqslant b \geqslant 4$, the solutions are
$(a, b)=(9,4),(6,5)$.
Upon verification, the only solution that forms a triangle is
$$
a=6, b=5, c=4 \text {. }
$$

In summary, there exist triangles $\triangle ABC$ with integer side lengths that satisfy the conditions, with side lengths of $3,7,8$ or $4,5,6$.