Answer
Dependent.
The solution set is $\{(2-3t\ ,\ 3-5t,\ t) \ |\ \ t\in \mathbb{R}\}$
Work Step by Step
Start with the augmented matrix, row reduce to
reduced row echelon form (Gauss-Jordan.)
$\left[\begin{array}{llll}
2 & -3 & -9 & -5\\
1 & 0 & 3 & 2\\
-3 & 1 & -4 & -3
\end{array}\right]\rightarrow\left(\begin{array}{l}
R_{1}\leftrightarrow R_{2}.\\
.\\
.
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & 0 & 3 & 2\\
2 & -3 & -9 & -5\\
-3 & 1 & -4 & -3
\end{array}\right]\rightarrow\left(\begin{array}{l}
.\\
-2R_{1}.\\
+3R_{1}.
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & 0 & 3 & 2\\
0 & -3 & -15 & -9\\
0 & 1 & 5 & 3
\end{array}\right]\rightarrow\left(\begin{array}{l}
.\\
\div(-3).\\
.
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & 0 & 3 & 2\\
0 & 1 & 5 & 3\\
0 & 1 & 5 & 3
\end{array}\right]\rightarrow\left(\begin{array}{l}
.\\
.\\
-R_{2}.
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & 0 & 3 & 2\\
0 & 1 & 5 & 3\\
0 & 0 & 0 & 0
\end{array}\right]$
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 $\Rightarrow y=3-5t$
Row $1 \Rightarrow x=2-3t$
The solution set is $\{(2-3t\ ,\ 3-5t,\ t) \ |\ \ t\in \mathbb{R}\}$
