Find the solution set to the corresponding homogeneous system of equations We are given a system of equations:
$2w+3x-2y+z=-1$
$6w+10x+6z=14$
$3w+2.5x-15y-4.5z=-35.5$
and a particular solution to that system of equations, $\begin{bmatrix}0\\2\\3\\-1\end{bmatrix}$
A) Find the solution set to the corresponding homogeneous system of equations, with all variables written in terms of free variables.
So my first question is do I set up the augmented matrix with each equation equal to $0$ to find the homogeneous solution or do I set it up how it is given? This is what I have of far but at this point I'm stuck...
$$\left(\begin{array} {cccc|c}
2 & 3 & -2 & 1 & 0\\
6 & 10 & 0 & 6 & 0\\
3 & 2.5 & -15 & -4.5 & 0\\
\end{array}\right)
\text{ }^{R2}\longleftrightarrow^{R3}
\left(\begin{array} {cccc|c}
2 & 3 & -2 & 1 & 0\\
3 & 2.5 & -15 & -4.5 & 0\\
6 & 10 & 0 & 6 & 0\\
\end{array}\right)$$
$$\xrightarrow{-2R_2+R_3\to R_3}
\left(\begin{array} {cccc|c}
2 & 3 & -2 & 1 & 0\\
3 & 2.5 & -15 & -4.5 & 0\\
0 & -5 & -30 & 15 & 0\\
\end{array}\right)\xrightarrow{2R_2+R_3\to R_2}
\left(\begin{array} {cccc|c}
2 & 3 & -2 & 1 & 0\\
3 & 2.5 & -15 & -4.5 & 0\\
0 & 0 & -60 & 6 & 0\\
\end{array}\right)$$
 A: (I edit my text for providing a solution closer to what you are looking for).
The fact that you give the general solution of the original system has no correlation with the question "solve the corresponding homogeneous system".
[In fact, it will be important to have both for an inevitable second question which is "deduce the  general solution of the homogeneous system". See last line of this answer]. 
Some keypoints:
a) Consider the issue as looking for the kernel of a $3 \times 4$ matrix $A$ which acts as a linear operator with source space $\mathbb{R}^4$ and range space $\mathbb{R}^3$.
b) Minor the dimension of the range space $dim(Im(A)) \geq 2$ because the first two columns of $A$ are independant.
c) Use the rank-nullity theorem:
$dim(Ker(A))+dim(Im(A)=dim(source \ space))=4$.
Thus $dim(Ker(A)) \leq 2$ and in fact, we are able to exhibit two independent vectors of the kernel (by looking for null linear combinations of the columns of $A$, by trial and error for example, which usually does not take a long time when the coefficients are integers or almost integers as is the case here. The coefficients of the null combinations are naturaly the coordinates of elements of the kernel).
One of the elements of the kernel is $(10,-6,1,0)$, another one, non proportional to the first, is $(0,3,2,-5)$ ; thus we have a basis of the kernel.
As a consequence, a final solution to your question is 
$(w,x,y,z)=a(10,-6,1,0)+b(0,3,2,-5)$ for any $a,b$.
Remark: this solution could appear under very different forms according to the basis chosen for $Ker(A)$.
and, consequently, the general solution is 
$(w,x,y,z)=a(10,-6,1,0)+b(0,3,2,-5)+(0,2,3,-1)$ for any $a,b$.
