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

Thus,
$$y z \leqslant 3$$

Therefore,
$$(y, z)=(1,1),(2,1),(3,1)$$

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
$$a b=a+b+3$$
$$(a-1)(b-1)=4$$
$$(a, b)=(5,2),(3,3)$$

When $(y, z)=(2,1)$, $2 a b c=2 x+6=a+b+c+6$, similar discussion shows that $c=1$, thus
$$(2 a-1)(2 b-1)=15$$

We get
$$(a, b)=(3,2)$$

When $(y, z)=(3,1)$, $3 a b c=3 x+12=a+b+c+12$, similarly, there is no solution in this case.

In summary, we have
$$\begin{aligned} & (a, b, c, x, y, z) \\ = & (2,2,2,6,1,1),(5,2,1,8,1,1),(3,3,1,7,1,1) \\ & (3,2,1,3,2,1),(6,1,1,2,2,2),(8,1,1,5,2,1) \\ & (7,1,1,3,3,1) \end{aligned}$$