6.214. $\left\{\begin{array}{l}x+y+z=0, \\ 2 x+3 y+z=0, \\ (x+1)^{2}+(y+2)^{2}+(z+3)^{2}=14 .\end{array}\right.$
## Solution.

From the first equation of the system, we find $z=-x-y$ and substitute it into the second and third equations:

$$
\begin{aligned}
& \left\{\begin{array}{l}
2 x+3 y-x-y=0 \\
(x+1)^{2}+(y+2)^{2}+(-x-y+3)^{2}=14
\end{array} \Leftrightarrow\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
x+2 y=0 \\
(x+1)^{2}+(y+2)^{2}+(-x-y+3)^{2}=14
\end{array}\right.
\end{aligned}
$$

From here, $x=-2 y,(-2 y+1)^{2}+(y+2)^{2}+(2 y-y+3)^{2}=14 \Leftrightarrow$ $\Leftrightarrow y^{2}+y=0$, from which $y_{1}=0, y_{2}=-1 ; x_{1}=0, x_{2}=2 ; z_{1}=0, z_{2}=-1$.'

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