Answer
$(-1,4,0)$
Work Step by Step
Start with the augmented matrix, row reduce to
reduced row echelon form (Gauss-Jordan.)
$\left[\begin{array}{llll}
1 & 1 & 6 & 3\\
1 & 1 & 3 & 3\\
1 & 2 & 4 & 7
\end{array}\right]\rightarrow\left(\begin{array}{l}
.\\
R_{3}\leftarrow R_{3}-R_{1}.\\
R_{3}\leftarrow R_{3}-R_{1}
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & 1 & 6 & 3\\
0 & 0 & -3 & 0\\
0 & 1 & -2 & 4
\end{array}\right]\rightarrow\left(\begin{array}{l}
.\\
R_{3}\leftarrow R_{2}.\\
.
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & 1 & 6 & 3\\
0 & 1 & -2 & 4\\
0 & 0 & -3 & 0
\end{array}\right]\rightarrow\left(\begin{array}{l}
R_{1}\leftarrow R_{1}-R_{2}.\\
.\\
\div(-3).
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & 0 & 8 & -1\\
0 & 1 & -2 & 4\\
0 & 0 & 1 & 0
\end{array}\right]\rightarrow\left(\begin{array}{l}
R_{1}\leftarrow R_{1}-8R_{3}.\\
R_{2}\leftarrow R_{2}+2R_{2}.\\
.
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & 0 & 0 & -1\\
0 & 1 & 0 & 4\\
0 & 0 & 1 & 0
\end{array}\right]$
The solution is $(-1,4,0)$