Answer
$\{(x,y,z)\ |\ x=\displaystyle \frac{2+4y}{7},\ z=\displaystyle \frac{4-13y}{7},\ y\in \mathbb{R}\}$
Work Step by Step
$\left\{\begin{array}{lllll}
2x & -3y & -z & =0 & \\
3x & +2y & +2z & =2 & /R_{2}=r_{2}+2r_{1}\\
x & +5y & +3z & =0 & /R_{3}=r_{3}+3r_{1}
\end{array}\right.$
... Eliminate $z$ in eq. 2 and 3
$\left\{\begin{array}{lllll}
2x & -3y & -z & =0 & \\
7x & -4y & & =2 & \\
7x & -4y & & =0 & /R_{3}=r_{3}-r_{2}
\end{array}\right.$
... Eliminate y in equation 3
$\left\{\begin{array}{lllll}
2x & -3y & -z & =0 & (1)\\
7x & -4y & & =2 & (2)\\
& & 0 & =0 & (3)
\end{array}\right.$
Equation 3 is 0=0, which is always satisfied.
The system is consistent and dependent.
Let $y$ be the parameter, $y\in \mathbb{R}.$
$(2)\displaystyle \Rightarrow 7x=2+4y\Rightarrow x=\frac{2+4y}{7}$
$(1)\displaystyle \Rightarrow z=2x-3y=\frac{2(2+4y)}{7}-\frac{21y}{7}=\frac{4-13y}{7}$
Solution set: $\{(x,y,z)\ |\ x=\displaystyle \frac{2+4y}{7},\ z=\displaystyle \frac{4-13y}{7},\ y\in \mathbb{R}\}$
