Answer
The solution set is $\{(r,s,t) \ |\ r=u, \ s=5, \ t=u,\ u\in \mathbb{R}\}$
(Dependent system)
Work Step by Step
Start with the augmented matrix, row reduce to
reduced row echelon form (Gauss-Jordan.)
$\left[\begin{array}{rrrr}
3 & 2 & -3 & 10\\
1 & -1 & -1 & -5\\
1 & 4 & -1 & 20
\end{array}\right]\rightarrow\left(\begin{array}{l}
R_{1}\leftrightarrow R_{2}.\\
.\\
-R_{2}.
\end{array}\right)$
$\rightarrow\left[\begin{array}{rrrr}
1 & -1 & -1 & -5\\
3 & 2 & -3 & 10\\
0 & 6 & 0 & 30
\end{array}\right]\rightarrow\left(\begin{array}{l}
.\\
-3R_{1}.\\
\div 6.
\end{array}\right)$
$\rightarrow\left[\begin{array}{rrrr}
1 & -1 & -1 & -5\\
0 & 5 & 0 & 25\\
0 & 1 & 0 & 5
\end{array}\right]\rightarrow\left(\begin{array}{l}
.\\
\div 5.\\
.
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & -1 & -1 & -5\\
0 & 1 & 0 & 5\\
0 & 1 & 0 & 5
\end{array}\right]\rightarrow\left(\begin{array}{l}
+R_{2}.\\
.\\
-R_{2}.
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & 0 & -1 & 0\\
0 & 1 & 0 & 5\\
0 & 0 & 0 & 0
\end{array}\right]$
The system is consistent (has solutions) and dependent.
$t$ has no leading 1 in its corresponding column.
Let $t=u\in \mathbb{R}$, an arbitrary number,
Row $2 \Rightarrow s=5$
Row $1 \Rightarrow r=u$
The solution set is $\{(r,s,t) \ |\ r=u, \ s=5, \ t=u,\ u\in \mathbb{R}\}$
(Dependent system)