Let us express the null space of $A$ as the span of a set of vectors, applying Theorem SSNS as economically as possible, without reference to the underlying homogeneous system of equations (in contrast to Example SSNS). \begin{equation*} A= \begin{bmatrix} 2 & 1 & 5 & 1 & 5 & 1 \\ 1 & 1 & 3 & 1 & 6 & -1 \\ -1 & 1 & -1 & 0 & 4 & -3 \\ -3 & 2 & -4 & -4 & -7 & 0 \\ 3 & -1 & 5 & 2 & 2 & 3 \end{bmatrix} \end{equation*}
Theorem SSNS creates vectors for the span by first row-reducing the matrix in question. The row-reduced version of $A$ is \begin{equation*} B= \begin{bmatrix} \leading{1} & 0 & 2 & 0 & -1 & 2 \\ 0 & \leading{1} & 1 & 0 & 3 & -1 \\ 0 & 0 & 0 & \leading{1} & 4 & -2 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} \end{equation*}
We will mechanically follow the prescription of Theorem SSNS. Here we go, in two big steps.
First, the non-pivot columns have indices $F=\set{3,\,5,\,6}$, so we will construct the $n-r=6-3=3$ vectors with a pattern of zeros and ones dictated by the indices in $F$. This is the realization of the first two lines of the three-case definition of the vectors $\vect{z}_j$, $1\leq j\leq n-r$. \begin{align*} \vect{z}_1&=\colvector{\\ \\ 1\\ \\0 \\ 0} & \vect{z}_2&=\colvector{\\ \\ 0\\ \\ 1\\ 0} & \vect{z}_3&=\colvector{\\ \\ 0\\ \\ 0\\ 1} \end{align*}
Each of these vectors arises due to the presence of a column that is not a pivot column. The remaining entries of each vector are the entries of the non-pivot column, negated, and distributed into the empty slots in order (these slots have indices in the set $D$, so also refer to pivot columns). This is the realization of the third line of the three-case definition of the vectors $\vect{z}_j$, $1\leq j\leq n-r$. \begin{align*} \vect{z}_1&=\colvector{-2\\ -1\\ 1\\ 0 \\0 \\ 0} & \vect{z}_2&=\colvector{1\\ -3\\ 0\\ -4\\ 1\\ 0} & \vect{z}_3&=\colvector{-2\\ 1\\ 0\\ 2\\ 0\\ 1} \end{align*}
So, by Theorem SSNS, we have \begin{equation*} \nsp{A} = \spn{\set{\vect{z}_1,\,\vect{z}_2,\,\vect{z}_3}} = \spn{\set{ \colvector{-2\\ -1\\ 1\\ 0 \\0 \\ 0},\, \colvector{1\\ -3\\ 0\\ -4\\ 1\\ 0},\, \colvector{-2\\ 1\\ 0\\ 2\\ 0\\ 1} }} \end{equation*}
We know that the null space of $A$ is the solution set of the homogeneous system $\homosystem{A}$, but nowhere in this application of Theorem SSNS have we found occasion to reference the variables or equations of this system. These details are all buried in the proof of Theorem SSNS.