Answer
No solution.
Work Step by Step
The given system of equations can be written in matrix form as below: $\left[ \left. \begin{matrix}
2 & -4 & 1 \\
1 & -3 & 1 \\
3 & -7 & 2 \\
\end{matrix} \right|\begin{matrix}
3 \\
5 \\
12 \\
\end{matrix} \right]$
Solve this matrix as below to get:
$\left[ \left. \begin{matrix}
2 & -4 & 1 \\
1 & -3 & 1 \\
0 & -2 & 1 \\
\end{matrix} \right|\begin{matrix}
3 \\
5 \\
3 \\
\end{matrix} \right]$ $ By,\ 3{{R}_{2}}-{{R}_{3}}\to {{R}_{3}}$
$\left[ \left. \begin{matrix}
2 & -4 & 1 \\
0 & -2 & 1 \\
0 & -2 & 1 \\
\end{matrix} \right|\begin{matrix}
3 \\
7 \\
3 \\
\end{matrix} \right]$ $ By,\ 2{{R}_{2}}-{{R}_{1}}\to {{R}_{2}}$
$\left[ \left. \begin{matrix}
2 & -4 & 1 \\
0 & -2 & 1 \\
0 & 0 & 0 \\
\end{matrix} \right|\begin{matrix}
3 \\
7 \\
4 \\
\end{matrix} \right]$ $ By,\ {{R}_{2}}-{{R}_{3}}\to {{R}_{3}}$
Convert the last row in equation form, $\begin{align}
0x+0y+0z=4 & \\
0=4 & \\
\end{align}$
And this statement is not true, so there is no solution of this system.
Hence, there is no solution for this system of equations.
