Let us determine a compact expression for the entire column space of the coefficient matrix of the system of equations that is Archetype D. Notice that in Example CSMCS we were only determining if individual vectors were in the column space or not, now we are describing the entire column space.
To start with the application of Theorem BCS, call the coefficient matrix $A$ and row-reduce it to reduced row-echelon form $B$, \begin{align*}A&=\begin{bmatrix} 2 & 1 & 7 & -7\\ -3 & 4 & -5 & -6\\ 1 & 1 & 4 & -5 \end{bmatrix} &B&=\begin{bmatrix} \leading{1} & 0 & 3 & -2\\ 0 & \leading{1} & 1 & -3\\ 0 & 0 & 0 & 0 \end{bmatrix} \end{align*}
Since columns 1 and 2 are pivot columns, $D=\{1,\,2\}$. To construct a set that spans $\csp{A}$, just grab the columns of $A$ with indices in $D$, so \begin{equation*} \csp{A}=\spn{\set{\colvector{2\\-3\\1},\,\colvector{1\\4\\1}}}. \end{equation*} That's it.
In Example CSMCS we determined that the vector \begin{equation*} \vect{c}=\colvector{2\\3\\2} \end{equation*} was not in the column space of $A$. Try to write $\vect{c}$ as a linear combination of the first two columns of $A$. What happens?
Also in Example CSMCS we determined that the vector \begin{equation*} \vect{b}=\colvector{8\\-12\\4} \end{equation*} was in the column space of $A$. Try to write $\vect{b}$ as a linear combination of the first two columns of $A$. What happens? Did you find a unique solution to this question? Hmmmm.