The number of solutions of the system of the system

The given system of equations is $x+2 y+z=3,2 x+3 y+z=3$ and $3 x+5 y+2 z=1$.

$\Delta=\left|\begin{array}{lll}1 & 2 & 1 \\ 2 & 3 & 1 \\ 3 & 5 & 2\end{array}\right|=1(6-5)-2(4-3)+1(4-3)=1-2+1=0$

$\Delta_{x}=\left|\begin{array}{lll}3 & 2 & 1 \\ 3 & 3 & 1 \\ 1 & 5 & 2\end{array}\right|=3(6-5)-2(6-1)+1(15-3)=3-10+12=5 \neq 0$

Here, $\Delta=0$ and at least one of $\Delta_{x}, \Delta_{y}$ and $\Delta_{z}$ is not zero, so the given system of equations has no solution.

Thus, the given system of equations has no solution.

The number of solutions of the system of the system of equations x + 2y + z = 3, 2x + 3y + z = 3, 3x + 5y + 2z = 1 is ___0___