If the system of linear equations $2x+2y+3z=a,3x-y+5z=b,x-3y+2z=c$ where $a,b,c$ are non-zero real numbers, has more than one solution, then

$A=\left[\begin{array}{ccc}2& 2& 3\\ 3& -1& 5\\ 1& -3& 2\end{array}\right],X=\left[\begin{array}{l}x\\ y\\ z\end{array}\right]andB=\left[\begin{array}{l}a\\ b\\ c\end{array}\right]$

Here, $\left|A\right|=0$
Since, the system of linear equations has more than one solution, so 

$\left(adjA\right)\left(B\right)=0\Rightarrow \left[\begin{array}{ccc}13& -13& 13\\ -1& 1& -1\\ -8& 8& -8\end{array}\right]\left[\begin{array}{l}a\\ b\\ c\end{array}\right]=\left[\begin{array}{l}0\\ 0\\ 0\end{array}\right]\Rightarrow b=a+c$