The definition of the column space is the span of the set of columns (Definition CSM). So the desired set is just the four columns of $B$,
\begin{equation*}
S=\set{
\colvector{2\\1\\-1},\,
\colvector{3\\1\\2},\,
\colvector{1\\0\\3},\,
\colvector{1\\1\\-4}
}
\end{equation*}
Theorem BCS suggests row-reducing the matrix and using the columns of $B$ with the same indices as the pivot columns.
\begin{equation*}
B\rref
\begin{bmatrix}
\leading{1} & 0 & -1 & 2\\
0 & \leading{1} & 1 & -1\\
0 & 0 & 0 & 0
\end{bmatrix}
\end{equation*}
So the pivot columns are numbered by elements of $D=\set{1,\,2}$, so the requested set is
\begin{equation*}
S=\set{
\colvector{2\\1\\-1},\,
\colvector{3\\1\\2}}
\end{equation*}
We can find this set by row-reducing the transpose of $B$, deleting the zero rows, and using the nonzero rows as column vectors in the set. This is an application of Theorem CSRST followed by Theorem BRS.
\begin{equation*}
\transpose{B}\rref
\begin{bmatrix}
\leading{1} & 0 & 3\\
0 & \leading{1} & -7\\
0 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}
\end{equation*}
So the requested set is
\begin{equation*}
S=\set{
\colvector{1\\0\\3},\,
\colvector{0\\1\\-7}
}
\end{equation*}
With the column space expressed as a null space, the vectors obtained via Theorem BNS will be of the desired shape. So we first proceed with Theorem FS and create the extended echelon form,
\begin{equation*}
\augmented{B}{I_3}=
\begin{bmatrix}
2 & 3 & 1 & 1 & 1 & 0 & 0\\
1 & 1 & 0 & 1 & 0 & 1 & 0\\
-1 & 2 & 3 & -4 & 0 & 0 & 1
\end{bmatrix}
\rref
\begin{bmatrix}
\leading{1} & 0 & -1 & 2 & 0 & \frac{2}{3} & \frac{-1}{3}\\
0 & \leading{1} & 1 & -1 & 0 & \frac{1}{3} & \frac{1}{3}\\
0 & 0 & 0 & 0 & \leading{1} & \frac{-7}{3} & \frac{-1}{3}
\end{bmatrix}
\end{equation*}
So, employing Theorem FS, we have $\csp{B}=\nsp{L}$, where
\begin{equation*}
L=
\begin{bmatrix}
\leading{1} & \frac{-7}{3} & \frac{-1}{3}
\end{bmatrix}
\end{equation*}
We can find the desired set of vectors from Theorem BNS as
\begin{equation*}
S=\set{
\colvector{\frac{7}{3}\\1\\0},\,
\colvector{\frac{1}{3}\\0\\1}
}
\end{equation*}