Answer
( $5-2t$ , $t-2$ , $t$ ),$\ \ t\in \mathbb{R}$
Work Step by Step
Write the augmented matrix and,
using row transformations,
arrive at the row-reduced echelon form.
$\left[\begin{array}{llll}
1 & 4 & -2 & -3\\
2 & -1 & 5 & 12\\
8 & 5 & 11 & 30
\end{array}\right] \ \ \begin{array}{l}
.\\
-2R_{1}.\\
-8R_{1}.
\end{array}$
$\left[\begin{array}{llll}
1 & 4 & -2 & -3\\
0 & -9 & 9 & 18\\
0 & -27 & 27 & 54
\end{array}\right] \ \ \begin{array}{l}
.\\
.\div(-9)\\
-3R_{2}.
\end{array}$
$\left[\begin{array}{llll}
1 & 4 & -2 & -3\\
0 & 1 & -1 & -2\\
0 & 0 & 0 & 0
\end{array}\right] \ \ \begin{array}{l}
-4R_{2}.\\
.\\
.\\
\end{array} $
$\left[\begin{array}{llll}
1 & 0 & 2 & 5\\
0 & 1 & -1 & -2\\
0 & 0 & 0 & 0
\end{array}\right]$
Last row has all zeros, so
the system is dependent (infinite number of solutions).
Taking $z=t\in \mathbb{R}$ (arbitrary),
back-substituting:
$x=5-2t$
$y=t-2$
Solution set: $\{$( $5-2t$ , $t-2$ , $t$ ),$\ \ t\in \mathbb{R} \}$
