Consider the column space of the $3\times 4$ matrix $A$, \begin{equation*} A= \begin{bmatrix} 3 & 2 & 1 & -4\\ -1 & 1 & -2 & 3\\ 2 & -4 & 6 & -8 \end{bmatrix} \end{equation*}
We first show that $\vect{v}=\colvector{18\\-6\\12}$ is in the column space of $A$, $\vect{v}\in\csp{A}$. Theorem CSCS says we need only check the consistency of $\linearsystem{A}{\vect{v}}$. Form the augmented matrix and row-reduce, \begin{equation*} \begin{bmatrix} 3 & 2 & 1 & -4 & 18\\ -1 & 1 & -2 & 3 & -6\\ 2 & -4 & 6 & -8 & 12 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & 0 & 1 & -2 & 6\\ 0 & \leading{1} & -1 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} \end{equation*}
Since the final column is not a pivot column, Theorem RCLS tells us the system is consistent and therefore by Theorem CSCS, $\vect{v}\in\csp{A}$.
If we wished to demonstrate explicitly that $\vect{v}$ is a linear combination of the columns of $A$, we can find a solution (any solution) of $\linearsystem{A}{\vect{v}}$ and use Theorem SLSLC to construct the desired linear combination. For example, set the free variables to $x_3=2$ and $x_4=1$. Then a solution has $x_2=1$ and $x_1=6$. Then by Theorem SLSLC, \begin{equation*} \vect{v}=\colvector{18\\-6\\12}= 6\colvector{3\\-1\\2}+ 1\colvector{2\\1\\-4}+ 2\colvector{1\\-2\\6}+ 1\colvector{-4\\3\\-8} \end{equation*}
Now we show that $\vect{w}=\colvector{2\\1\\-3}$ is not in the column space of $A$, $\vect{w}\not\in\csp{A}$. Theorem CSCS says we need only check the consistency of $\linearsystem{A}{\vect{w}}$. Form the augmented matrix and row-reduce, \begin{equation*} \begin{bmatrix} 3 & 2 & 1 & -4 & 2\\ -1 & 1 & -2 & 3 & 1\\ 2 & -4 & 6 & -8 & -3 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & 0 & 1 & -2 & 0\\ 0 & \leading{1} & -1 & 1 & 0\\ 0 & 0 & 0 & 0 & \leading{1} \end{bmatrix} \end{equation*}
since the final column is a pivot column, Theorem RCLS tells us the system is inconsistent and therefore by Theorem CSCS, $\vect{w}\not\in\csp{A}$.