6.221. $\left\{\begin{array}{l}x+y+z=2, \\ 2 x+3 y+z=1, \\ x^{2}+(y+2)^{2}+(z-1)^{2}=9 .\end{array}\right.$
## Solution.

From the first equation of the system, we find $z=2-x-y$.

$$
\left\{\begin{array} { l } 
{ 2 x + 3 y + 2 - x - y = 1 , } \\
{ x ^ { 2 } + ( y + 2 ) ^ { 2 } + ( 2 - x - y - 1 ) ^ { 2 } = 9 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x+2 y=-1, \\
x^{2}+(y+2)^{2}+(1-x-y)^{2}=9
\end{array} \Rightarrow\right.\right.
$$

$x=-1-2 y,(-1-2 y)^{2}+(y+2)^{2}+(1+1+2 y-y)^{2}=9 \Leftrightarrow 6 y^{2}+12 y=0$, from which $y_{1}=0, y_{2}=-2$; then $x_{1}=-1, x_{2}=3 ; z_{1}=3, z_{2}=1$.

Answer: $(-1 ; 0 ; 3),(3 ;-2 ; 1)$.