## System of equations problem

Hi MHF,
Here is an exercise and I've done part a) and b) but not part c). So I'd like a check up for the 2 first parts and a tip for the last part. Thanks in advance.
Consider the system of equations given by $w-2x-y+3z=a$, $-w+x-2y+2z=b$, $2w-3x+y+z=c$.
a)Show that the system has at least 2 solutions that are linear independent if $a=b=c=0$.
My attempt : I wrote the system of equations as a matrix and reduced it in order to get $\begin{bmatrix} 1 & 0 & 5 & -7 \\ 0 & 1 & 3 & -5 \\ 0 & 0 & 0 & 0 \end{bmatrix}$. From which we get $\begin{cases} w+5y-7z=0 \\ x+3y-5z=0 \end{cases}$ so any vector in the plane can be written as $y(-5, -3, 1, 0)+z(7, 5, 0, 1)$. Let $y=1$ and $z=0$, we get that a solution is $\{ (-5, -3, 1, 0 ) \}$, let $y=0$ and $z=1$ and we get $\{ (7, 5, 0, 1 ) \}$ which are clearly 2 linear independent solutions.

b)Determine the values of a, b and c such that the system has solution.
My attempt : I wrote the system as an augmented matrix (augmented with the column a, b, c) and finally reaches the same matrix as before of course, but with the column vector $(2a+b, -b-a, c-2a+b+a)$ from which we have $\begin{cases} w+5y-7z=2a+b \\ x+3y-5z=-b-a \\ 0=c-2a+b+a \end{cases}$ so $w+x+8y-12z=a$, $b$ is any value and $c=a-b$.

c)Show all the solutions for $a=1$, $b=0$ and $c=1$, if there are/is.
My attempt : $\begin{cases} w+5y-7z=2 \\ x+3y-5z=-1 \end{cases}\Rightarrow w+x+8y-12z=1$ so the solutions are infinitely many? All numbers $w$, $x$, $y$ and $z \in \mathbb{R}$ such that $w+x+8y-12z=1$ are solutions?