Begin with the finite set of three vectors of size $3$ that are the columns of the coefficient matrix in Archetype B, \begin{equation*} R=\{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3\} =\left\{ \colvector{-7\\5\\1},\,\colvector{-6\\5\\0},\,\colvector{-12\\7\\4} \right\} \end{equation*} and consider the infinite set $\spn{R}$.
First, as an example, note that \begin{equation*} \vect{x}=(2)\colvector{-7\\5\\1}+(4)\colvector{-6\\5\\0}+(-3)\colvector{-12\\7\\4}= \colvector{-2\\9\\-10} \end{equation*} is in $\spn{R}$, since it is a linear combination of $\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3$. In other words, $\vect{x}\in\spn{R}$. Try some different values of $\alpha_1,\,\alpha_2,\,\alpha_3$ yourself, and see what vectors you can create as elements of $\spn{R}$.
Now ask if a given vector is an element of $\spn{R}$. For example, is $\vect{z}=\colvector{-33\\24\\5}$ in $\spn{R}$? Is $\vect{z}\in\spn{R}$?
To answer this question, we will look for scalars $\alpha_1,\,\alpha_2,\,\alpha_3$ so that \begin{align*} \alpha_1\vect{v}_1+\alpha_2\vect{v}_2+\alpha_3\vect{v}_3&=\vect{z}\\ \end{align*} By Theorem SLSLC solutions to this vector equation are the solutions to the system of equations \begin{align*} -7\alpha_1-6\alpha_2-12\alpha_3&=-33\\ 5\alpha_1+5\alpha_2+7\alpha_3&=24\\ \alpha_1+4\alpha_3&=5 \end{align*} Building the augmented matrix for this linear system, and row-reducing, gives \begin{align*} \begin{bmatrix} \leading{1}&0&0&-3\\ 0&\leading{1}&0&5\\ 0&0&\leading{1}&2 \end{bmatrix} \end{align*}
This system has a unique solution, \begin{align*} \alpha_1 = -3&&\alpha_2 = 5&&\alpha_3 = 2 \end{align*} telling us that \begin{equation*} (-3)\vect{v}_1+(5)\vect{v}_2+(2)\vect{v}_3=\vect{z} \end{equation*} so we are convinced that $\vect{z}$ really is in $\spn{R}$. Notice that in this case we have only one way to answer the question affirmatively since the solution is unique.
Let us ask about another vector, say is $\vect{x}=\colvector{-7\\8\\-3}$ in $\spn{R}$? Is $\vect{x}\in\spn{R}$?
We desire scalars $\alpha_1,\,\alpha_2,\,\alpha_3$ so that \begin{align*} \alpha_1\vect{v}_1+\alpha_2\vect{v}_2+\alpha_3\vect{v}_3&=\vect{x}\\ \end{align*} By Theorem SLSLC solutions to this vector equation are the solutions to the system of equations \begin{align*} -7\alpha_1-6\alpha_2-12\alpha_3&=-7\\ 5\alpha_1+5\alpha_2+7\alpha_3&=8\\ \alpha_1+4\alpha_3&=-3 \end{align*} Building the augmented matrix for this linear system, and row-reducing, gives \begin{align*} \begin{bmatrix} \leading{1} & 0 & 0 & 1 \\ 0 & \leading{1} & 0 & 2 \\ 0 & 0 & \leading{1} & -1 \end{bmatrix} \end{align*}
This system has a unique solution, \begin{align*} \alpha_1 = 1&&\alpha_2 = 2&&\alpha_3 = -1 \end{align*} telling us that \begin{equation*} (1)\vect{v}_1+(2)\vect{v}_2+(-1)\vect{v}_3=\vect{x} \end{equation*} so we are convinced that $\vect{x}$ really is in $\spn{R}$. Notice that in this case we again have only one way to answer the question affirmatively since the solution is again unique.
We could continue to test other vectors for membership in $\spn{R}$, but there is no point. A question about membership in $\spn{R}$ inevitably leads to a system of three equations in the three variables $\alpha_1,\,\alpha_2,\,\alpha_3$ with a coefficient matrix whose columns are the vectors $\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3$. This particular coefficient matrix is nonsingular, so by Theorem NMUS, the system is guaranteed to have a solution. (This solution is unique, but that is not critical here.) So no matter which vector we might have chosen for $\vect{z}$, we would have been certain to discover that it was an element of $\spn{R}$. Stated differently, every vector of size 3 is in $\spn{R}$, or $\spn{R}=\complex{3}$.
Compare this example with Example SCAA, and see if you can connect $\vect{z}$ with some aspects of the write-up for Archetype B.