The coefficient matrix in Archetype A is $A$, which row-reduces to $B$, \begin{align*}A&=\begin{bmatrix} 1 & -1 & 2\\ 2 & 1 & 1\\ 1 & 1 & 0 \end{bmatrix} &B&=\begin{bmatrix} \leading{1} & 0 & 1\\ 0 & \leading{1} & -1\\ 0 & 0 & 0 \end{bmatrix} \end{align*}
Columns 1 and 2 are pivot columns, so by Theorem BCS we can write \begin{equation*} \csp{A}=\spn{\set{\vect{A}_1,\,\vect{A}_2}}=\spn{\set{\colvector{1\\2\\1},\,\colvector{-1\\1\\1}}}. \end{equation*}
We want to show in this example that $\csp{A}\neq\complex{3}$. So take, for example, the vector $\vect{b}=\colvector{1\\3\\2}$. Then there is no solution to the system $\linearsystem{A}{\vect{b}}$, or equivalently, it is not possible to write $\vect{b}$ as a linear combination of $\vect{A}_1$ and $\vect{A}_2$. Try one of these two computations yourself. (Or try both!). Since $\vect{b}\not\in\csp{A}$, the column space of $A$ cannot be all of $\complex{3}$. So by varying the vector of constants, it is possible to create inconsistent systems of equations with this coefficient matrix (the vector $\vect{b}$ being one such example).
In Example MWIAA we wished to show that the coefficient matrix from Archetype A was not invertible as a first example of a matrix without an inverse. Our device there was to find an inconsistent linear system with $A$ as the coefficient matrix. The vector of constants in that example was $\vect{b}$, deliberately chosen outside the column space of $A$.