6.095. $\left\{\begin{array}{l}3 x+2 y+2 z=13, \\ 2 x+3 y+2 z=14, \\ 2 x+2 y+3 z=15 .\end{array}\right.$
## Solution.

By adding all three equations, we get $7(x+y+z)=42$, from which $x+y+z=6$. Now we will sequentially subtract this equation from each equation in the system:

$$
\begin{aligned}
& \left\{\begin{array}{l}
3 x+2 y+2 z=13, \\
x+y+z=6
\end{array} \Leftrightarrow x=1 ;\left\{\begin{array}{l}
2 x+3 y+2 z=14 \\
x+y+z=6
\end{array} \Leftrightarrow y=2 ;\right.\right. \\
& \left\{\begin{array}{l}
2 x+2 y+3 z=15, \\
x+y+z=6
\end{array} \Leftrightarrow z=3 .\right.
\end{aligned}
$$

Answer: $(1 ; 2 ; 3)$.