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?