Answer
The system has no solution.
Work Step by Step
Start with the augmented matrix.
Use matrix row operations to get ls down the main diagonal from upper left to lower right, and 0s below the $1\mathrm{s}$ (row-echelon form).
$\left[\begin{array}{lllll}
2 & -4 & 1 & | & 3\\
1 & -3 & 1 & | & 5\\
3 & -7 & 2 & | & 12
\end{array}\right]$
Swapping R1 and R2 brings a 1 into the first diagonal position
$\left[\begin{array}{lllll}
1 & -3 & 1 & | & 5\\
2 & -4 & 1 & | & 3\\
3 & -7 & 2 & | & 12
\end{array}\right]\left\{\begin{array}{l}
.\\
-2R1+R2\rightarrow R2.\\
-3R1+R3\rightarrow R3.
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & -3 & 1 & | & 5\\
0 & 2 & -1 & | & -7\\
0 & 2 & -1 & | & -3
\end{array}\right]\left\{\begin{array}{l}
.\\
.\\
-R2+R3\rightarrow R3.
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & -3 & 1 & | & 5\\
0 & 2 & -1 & | & -7\\
0 & 0 & 0 & | & 4
\end{array}\right]$
The last row represents the equation
0=4
which is impossible.
The system has no solution.
