865*. Solve the equation in natural numbers $x, y$ and $z$

$$
x y + y z + z x = 2(x + y + z).
$$
$\triangle$ Let's gather all the terms of the equation on the left side and rewrite it as:

$$
x(y-2)+y(z-2)+z(x-2)=0 .
$$

From the last equation, it follows that the inequalities

$$
x>2, \quad y>2, \quad z>2
$$

cannot all be true simultaneously, otherwise the left side of the equation would become positive. Therefore, at least one of the unknowns does not exceed 2. Suppose this is $x$. Let's consider two possibilities.

1) Let $x=1$.

Then the original equation becomes:

$$
y+y z+z=2+2 y+2 z, \quad y z=y+z+2
$$

Express $z$ in terms of $y$.

$$
y z-z=y+2, \quad z=\frac{y+2}{y-1}=\frac{(y-1)+3}{y-1}=1+\frac{3}{y-1} .
$$

Thus, $y-1=1$ or $y-1=3$. In the first case, $y=2, z=4$, in the second case, $y=4, z=2$.

2) Let $x=2$.

We get:

$$
2 y+y z+2 z=4+2 y+2 z, \quad y z=4
$$

From this, $y=1, z=4 ; y=4, z=1 ; y=z=2$.

Answer: $(1 ; 2 ; 4),(1 ; 4 ; 2),(2 ; 1 ; 4),(2 ; 4 ; 1),(2 ; 2 ; 2),(4 ; 1 ; 2),(4 ; 2 ; 1)$.