6.219. $\left\{\begin{array}{l}x+y+z=3, \\ x+2 y-z=2, \\ x+y z+z x=3 .\end{array}\right.$
## Solution.

From the first equation of the system, we find $z=3-x-y$. Then we get

$$
\begin{aligned}
& \left\{\begin{array}{l}
x+2 y-3+x+y=2 \\
x+y(3-x-y)+(3-x-y) x=3, \\
x+y z+z x=3
\end{array}\right. \\
& x+\frac{5-2 x}{3}\left(3-x-\frac{5-2 x}{3}\right)+\left(3-x-\frac{5-2 x}{3}\right.
\end{aligned}
$$

from which $x_{1}=1, x_{2}=7$. Then $y_{1}=1, y_{2}=-3 ; z_{1}=1, z_{2}=-1$.

Answer: $(1 ; 1 ; 1),(7 ;-3 ;-1)$.