Example 2 Find all integer solutions $(a, b, c, x, y, z)$ satisfying:
( i ) $\left\{\begin{array}{l}a+b+c=xyz, \\ x+y+z=abc\end{array}\right.$ ( ii ) $a \geqslant b \geqslant c \geqslant 1$, $x \geqslant y \geqslant z \geqslant 1$
Solution: By symmetry, we only consider the case $x \geqslant a$.
$\because x y z=a+b+c \leqslant 3 a \leqslant 3 x$
$\therefore y z \leqslant 3,(y, z)=(1,1),(2,1),(3,1)$.
(1) When $(y, z)=(1,1)$, $\left\{\begin{array}{l}a+b+c=x, \\ x+2=a b c\end{array}\right.$,
$a+b+c+2=a b c$.
If $c \geqslant 2$, then $a+b+c+2 \leqslant 3 a+2 \leqslant 4 a \leqslant$ $a b c$, where the equality holds only when $a=b=c=2$, in this case $x=6$;
If $c=1$, then $a+b+3=a b$,
i.e., $(a-1)(b-1)=4$,
$\therefore(a, b)=(5,2),(3,3), x=7,8$.
(2) When $(y, z)=(2,1)$, $\left\{\begin{array}{l}a+b+c=2 x, \\ x+3=a b c,\end{array} \Rightarrow a\right.$ $+b+c+6=2 a b c$.

If $c \geqslant 2$, then $a+b+c+6 \leqslant 3 a+6 \leqslant 6 a<8 a$ $\leqslant 2 a b c$, only $c=1$,

At this time $a+b+7=2 a b$, i.e., $(2 a-1)(2 b-1)=$ 15,
$$
\therefore\left\{\begin{array}{l}
2 a - 1 = 15, \\
2 b - 1 = 1,
\end{array} \text { or } \left\{\begin{array}{l}
2 a-1=5, \\
2 b-1=3 .
\end{array}(a, b)=\right.\right.
$$
$(3,2)((8,1)$ is discarded because $x=5<a)$.
(3) When $(y, z)=(3,1)$, $\left\{\begin{array}{l}a+b+c=3 x, \\ x+4=a b c,\end{array} \Rightarrow a\right.$ $+b+c+12=3 a b c$.
$\because 3 a \geqslant a+b+c=3 x \geqslant 3 a$,
$\therefore x=a,\left\{\begin{array}{l}b+c=2 a, \\ a+4=a b c\end{array}\right.$
$\because 2 a=b+c \leqslant 2 a$, the equality holds only when $a=b=c$, $\therefore a+4=a^{3}$, i.e., $a\left(a^{2}-1\right)=4$, no solution.
In summary, the integer solutions $(a, b, c, x, y, z)=(2$,
$$
\begin{array}{l}
2,2,6,1,1),(5,2,1,8,1,1),(3,3,1,7,1,1), \\
(3,2,1,3,2,1),(6,1,1,2,2,2),(8,1,1,5,2, \\
1),(7,1,1,3,3,1) .
\end{array}
$$