By symmetry, we only need to consider the case $x \geqslant a$. At this time, $x y z=a+b+c \leqslant 3 a \leqslant 3 x$, so $y z \leqslant 3$. Thus, $(y, z)=(1,1),(2,1),(3,1)$.

When $(y, z)=(1,1)$, $a+b+c=x$ and $x+2=a b c$, thus $a b c=a+b+c+2$. If $c \geqslant 2$, then
$$a+b+c+2 \leqslant 3 a+2 \leqslant 4 a \leqslant a b c$$

Equality holds if and only if $a=b=c=2$.
If $c=1$, then $a b=a+b+3$, which means
$$(a-1)(b-1)=4$$

Thus, $(a, b)=(5,2)$ or $(3,3)$.
When $(y, z)=(2,1)$,
$$2 a b c=2 x+6=a+b+c+6,$$

Similar discussion shows that $c=1$, then $(2 a-1)(2 b-1)=15$, yielding $(a, b)=(3, 2)$.

When $(y, z)=(3,1)$,
$$3 a b c=3 x+12=a+b+c+12,$$

Discussion shows that there is no solution for $x \geqslant a$.
In summary, we have $(a, b, c, x, y, z)=(2,2,2,6,1,1),(5,2,1,8,1,1)$, $(3,3,1,7,1,1),(3,2,1,3,2,1),(6,1,1,2,2,2),(8,1,1,5,2,1)$ and $(7, 1,1,3,3,1)$.
Example 5 Find all integer arrays $(a, b, c, x, y, z)$, such that
$$\left\{\begin{array}{l}
a+b+c=x y z, \\
x+y+z=a b c,
\end{array}\right.$$

where $a \geqslant b \geqslant c \geqslant 1, x \geqslant y \geqslant z \geqslant 1$.