Answer
The solution set is $\{(10-4t\ ,\ 3.5+0.5t,\ t) \ |\ \ t\in \mathbb{R}\}$
(Dependent system)
Work Step by Step
Start with the augmented matrix, row reduce to
reduced row echelon form (Gauss-Jordan.)
$\left[\begin{array}{rrrr}
1 & -2 & 5 & 3\\
-2 & 6 & -11 & 1\\
3 & -16 & 20 & -26
\end{array}\right]\rightarrow\left(\begin{array}{l}
.\\
+2R_{1}.\\
-3R_{1}.
\end{array}\right)$
$\rightarrow\left[\begin{array}{rrrr}
1 & -2 & 5 & 3\\
0 & 2 & -1 & 7\\
0 & -10 & 5 & -35
\end{array}\right]\rightarrow\left(\begin{array}{l}
+R_{2}.\\
\div 2.\\
+5R_{2}.
\end{array}\right)$
$\rightarrow\left[\begin{array}{rrrr}
1 & 0 & 4 & 10\\
0 & 1 & -0.5 & 3.5\\
0 & 0 & 0 & 0
\end{array}\right]$
The last row represents 0=0, which is always true.
The system is consistent (has solutions) and dependent.
Let $z=t\in \mathbb{R}$, an arbitrary number,
Row 2 $\Rightarrow y=3.5+0.5t$
Row $1 \Rightarrow x=10-4t$
The solution set is $\{(10-4t\ ,\ 3.5+0.5t,\ t) \ |\ \ t\in \mathbb{R}\}$
(Dependent system)