Answer
$\left[\begin{array}{l}
2\\
-1\\
-4\\
3
\end{array}\right]\in$ Col A, $\left[\begin{array}{l}
3\\
1
\end{array}\right]\in$Nul A.
Work Step by Step
A is a 4$\times$2 matrix.
Nul A is a subspace of $\mathbb{R}^{2}$.
Col A is a subspace of $\mathbb{R}^{4}$.
Col A= Span $\{ \left[\begin{array}{l}
2\\
-1\\
-4\\
3
\end{array}\right],\ \left[\begin{array}{l}
-6\\
3\\
12\\
-9
\end{array}\right] ]$,
$ \left[\begin{array}{l}
2\\
-1\\
-4\\
3
\end{array}\right]\in$ Col A.
For Nul A, find the general solution of Ax=0:
[A 0]=$\left[\begin{array}{lll}
2 & -6 & 0\\
-1 & 3 & 0\\
-4 & 12 & 0\\
3 & -9 & 0
\end{array}\right]\left[\begin{array}{l}
\div 2\\
.\\
.\\
.
\end{array}\right.$
$\sim\left[\begin{array}{lll}
1 & -3 & 0\\
-1 & 3 & 0\\
-4 & 12 & 0\\
3 & -9 & 0
\end{array}\right]\left[\begin{array}{l}
.\\
+R_{1}.\\
+4R_{1}.\\
-3R_{1}.
\end{array}\right.$
$\sim\left[\begin{array}{lll}
1 & -3 & 0\\
0 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}\right]$
With $x_{2}$ a free parameter, say $x_{2}=1$
$x_{1}=3x_{2}=3$
$\left[\begin{array}{l}
3\\
1
\end{array}\right]\in$Nul A.