Answer
The solution set is $\{(5-2t\ ,\ t-2,\ t) \ |\ \ t\in \mathbb{R}\}$
(Dependent system)
Work Step by Step
Start with the augmented matrix, row reduce to
reduced row echelon form (Gauss-Jordan.)
$\left[\begin{array}{llll}
1 & 4 & -2 & -3\\
2 & -1 & 5 & 12\\
8 & 5 & 11 & 30
\end{array}\right]\rightarrow\left(\begin{array}{l}
.\\
-2R_{1}.\\
-8R_{1}.
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & 4 & -2 & -3\\
0 & -9 & 9 & 18\\
0 & -27 & 27 & 54
\end{array}\right]\rightarrow\left(\begin{array}{l}
.\\
\div(-9.)\\
-3R_{2}.
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & 4 & -2 & -3\\
0 & 1 & -1 & -2\\
0 & 0 & 0 & 0
\end{array}\right]\rightarrow\left(\begin{array}{l}
-4R_{2}.\\
.\\
.
\end{array}\right)$
(the last row poses no problem, as it represents 0=0, which is always true)
$\rightarrow\left[\begin{array}{llll}
1 & 0 & 2 & 5\\
0 & 1 & -1 & -2\\
0 & 0 & 0 & 0
\end{array}\right]$
The system is consistent (has solutions) and dependent.
$z$ has no leading 1 in its corresponding column.
Let $z=t\in \mathbb{R}$, an arbitrary number,
Row $2 \Rightarrow y=t-2$
Row $1 \Rightarrow x=5-2t$
The solution set is $\{(5-2t\ ,\ t-2,\ t) \ |\ \ t\in \mathbb{R}\}$
(Dependent system)