6.097. $\left\{\begin{array}{l}x+2 y+3 z=3, \\ 3 x+y+2 z=7, \\ 2 x+3 y+z=2 .\end{array}\right.$
## Solution.

We will transform the system using the Gaussian method, i.e., subtract the first equation, multiplied by the corresponding number, from the second and third equations.

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ x + 2 y + 3 z = 3 , } \\
{ 3 x + y + 2 z = 7 , } \\
{ 2 x + 3 y + z = 2 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ x + 2 y + 3 z = 3 , } \\
{ - 5 y - 7 z = - 2 , } \\
{ - y - 5 z = - 4 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x+2 y+3 z=3, \\
y+5 z=4, \\
-5 y-7 z=-2
\end{array}\right.\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ x + 2 y + 3 z = 3 , } \\
{ y + 5 z = 4 , } \\
{ 1 8 z = 1 8 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ x = 3 - 2 y - 3 z , } \\
{ y = 4 - 5 z , } \\
{ z = 1 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=2, \\
y=-1, \\
z=1 .
\end{array}\right.\right.\right.
\end{aligned}
$$

Answer: $(2, -1, 1)$