Answer
( $ t$ , $5 ,\ t$ ),$\ \ t\in \mathbb{R}$
Work Step by Step
Write the augmented matrix and,
using row transformations,
arrive at the row-reduced echelon form.
$\left[\begin{array}{llll}
3 & 2 & -3 & 10\\
1 & -1 & -1 & -5\\
1 & 4 & -1 & 20
\end{array}\right]\ \ \begin{array}{l}
R_{1}\leftrightarrow R_{2}.\\
.\\
.
\end{array}$
$\left[\begin{array}{llll}
1 & -1 & -1 & -5\\
3 & 2 & -3 & 10\\
1 & 4 & -1 & 20
\end{array}\right]\ \ \begin{array}{l}
.\\
-3R_{1}.\\
-R_{1}.
\end{array}$
$\left[\begin{array}{llll}
1 & -1 & -1 & -5\\
0 & 5 & 0 & 25\\
0 & 5 & 0 & 25
\end{array}\right]\ \ \begin{array}{l}
.\\
\div 5.\\
-R_{2}.
\end{array}$
$\left[\begin{array}{llll}
1 & -1 & -1 & -5\\
0 & 1 & 0 & 5\\
0 & 0 & 0 & 0
\end{array}\right]\ \ \begin{array}{l}
+R_{2}.\\
.\\
.
\end{array}$
$\left[\begin{array}{llll}
1 & 0 & -1 & 0\\
0 & 1 & 0 & 5\\
0 & 0 & 0 & 0
\end{array}\right]$
Last row has all zeros, so
the system is dependent (infinite number of solutions).
Taking $t\in \mathbb{R}$ (arbitrary),
back-substituting:
$r=t$
$s=5$
Solution set: $\{$( $ t$ , $5 ,\ t$ ),$\ \ t\in \mathbb{R} \}$
