You have four equations and four unknowns. If this system of equations has a solution, you have everything you need to solve it. One thing I learned while taking algebra is to consider the system as a matrix. First set each equation equal to zero.
$$0a+3b-2c-0d=0 \\ 2a-1b+0c-2d=0 \\ -3a+0b+1c+3d = 0 \\ 0a-3b +2c+0d = 0$$
The next part can easily be calculated using a graphing calculator. Make a $4\times 5$ and fill it out with the coefficients of your equation above. It should look like this:
$$\begin{array}{columns}
0 & 3 & -2 & 0 & 0\\
2 & -1 & 0 & -2 & 0\\
-3 & 0 & 1 & 3 & 0\\
0 & -3 & 2 & 0& 0\\
\end{array}$$
Now you want to reduce the matrix to "Row reduced echelon form" which will ideally convert the matrix into something that looks like
$$\begin{array}{columns}
1 & 0 & 0 & 0 & x_1\\
0 & 1 & 0 & 0 & x_2\\
0 & 0 & 1 & 0 & x_3\\
0 & 0 & 0 & 1 & x_4\\
\end{array}$$
where $a=x_1, \space b = x_2, \space c=x_3$ and $d = x_4$ is your final answer. There should be a button under matrix tools on the calculator that is titled "rref" and you want to calculate $\text{rref}(M)$ where $M$ is the coefficient matrix above. If your matrix cannot be completely reduced into a diagonal of $1's$ with a single column of numbers on the far right, then there is not a unique solution to the system. There might be no solutions, or there could be infinite. At any rate, knowing how to do this on a calculator is extremely useful and can save a lot of time!