#### Answer

Inconsistent.

#### Work Step by Step

Write the augmented matrix of the system of linear equations.
$ \begin{bmatrix}
1 & 2 & 6 & 1\\
2 & 5 & 15 & 4\\
3 & 1 & 3 & -6
\end{bmatrix} $
Add -2 times the 1st row to the 2nd row to produce a new 2nd row.
Add -3 times the 1st row to the 3rd row to produce a new 3rd row.
$ \begin{bmatrix}
1 & 2 & 6 & 1\\
0 & 1 & 3 & 2\\
0& -5 & -15 & -9
\end{bmatrix} $
Add 5 times the 2nd row to the 3rd row to produce a new 3rd row.
$ \begin{bmatrix}
1 & 2 & 6 & 1\\
0 & 1 & 3 & 2\\
0& 0 & 0 & 1
\end{bmatrix} $
Use back-substitution to find the solution.
$0z=1 \rightarrow $ this equation is not possible $\rightarrow$ the system has no solution.