#### Answer

Infinitely many solutions; dependent equations.

#### Work Step by Step

The given system of equations is
$x-3y+z=2$
$4x-12y+4z=8$
$-2x+6y-2z=-4$
The augmented matrix is
$\Rightarrow \left[\begin{array}{ccc|c}
1 & -3 & 1& 2\\
4 & -12 & 4& 8 \\
-2&6&-2&-4
\end{array}\right]$
Perform $R_2\rightarrow R_2-4\times R_1$ and $R_3\rightarrow R_3+ 2R_1$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & -3 & 1& 2\\
4-4(1) & -12-4(-3) & 4-4(1) & 8-4(2) \\
-2+2(1)&6+2(-3)&-2+2(1)&-4+2(2)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & -3 & 1& 2\\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{array}\right]$
Use back substitution in the second and third row.
$\Rightarrow x(0)+y(0)+z(0)=0$
$\Rightarrow 0=0$.
Hence, the system of linear equation contains dependent equations and has infinitely many solutions.