Answer
$(x,-x+3,0)$
Work Step by Step
We have to solve the system:
$$\begin{align*}
\begin{cases}
x+y+z&=3\quad\quad\text{Equation }1\\
x+y-z&=3\quad\quad\text{Equation }2\\
2x+2y+z&=6\quad\quad\text{Equation }3.
\end{cases}
\end{align*}$$
We rewrite the system as a linear system in $\textit{two}$ variables::
$$\begin{align*}
x+y+z&=3\quad\quad\text{Add Equation }1\\
x+y-z&=3\quad\quad\text{to Equation }2\\
\text{___________}&\text{______}\\
2x+2y&=6\quad\quad\text{New Equation }1.
\end{align*}$$
$$\begin{align*}
x+y-z&=3\quad\quad\text{Add Equation }2\\
2x+2y+z&=6\quad\quad\text{to Equation }3\\
\text{___________}&\text{______}\\
3x+3y&=9\quad\quad\text{New Equation }2.
\end{align*}$$
We solve the new linear system for both its variables:
$$\begin{align*}
-6x-6y&=-18\quad\text{Add }-3\text{ times Equation }1\\
6x+6y&=18\quad\quad\text{to Equation }3\\
\text{___________}&\text{______}\\
0&=0.
\end{align*}$$
As we obtain the identity $0=0$, we conclude that the original system has infinitely many solutions.
Now we describe the solutions of the system. We divide new Equation $1$ by $2$ to get $x+y=3$ or $y=-x+3$. We substitute this into original Equation $1$ and obtain $z=0$.
So any ordered triple of the form $(x,-x+3,0)$ is a solution of the system.
