Answer
No solution or $\varnothing$.
Work Step by Step
The given system of equations is
$x-y+3z=4$
$2x-2y+6z=7$
$3x-y+5z=14$
The augmented matrix is
$\Rightarrow \left[\begin{array}{ccc|c}
1 & -1 & 3& 4\\
2 & -2 & 6& 7 \\
3&-1&5&14
\end{array}\right]$
Perform $R_2\rightarrow R_2-2R_1$ and $R_3\rightarrow R_3-3 R_1$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & -1 & 3& 4\\
2-2(1) & -2-2(-1) & 6-2(3)& 7-2(4) \\
3-3(1)&-1-3(-1)&5-3(3)&14-3(4)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & -1 & 3& 4\\
0 & 0 & 0& -1 \\
0&2&-4&2
\end{array}\right]$
Use back substitution for the second row.
$x(0)+y(0)+z(0)=-1$
There are no values of $x,y$ and $z$ for which the above equation satisfy.
Hence, the system is inconsistent and has no solution.
