Answer
$ w\in$Col A
and
$ w\in$Nul A.
Work Step by Step
A typical vector v in Nul A has the property that Av=0.
A typical vector v in Col A has the property that the equation Ax=v is consistent.
Is $Ax=w$ consistent$?$
[A w]=$\left[\begin{array}{llll}
-8 & -2 & -9 & 2\\
6 & 4 & 8 & 1\\
4 & 0 & 4 & -2
\end{array}\right]\left[\begin{array}{l}
R_{1}\leftrightarrow R_{3}.\\
.\\
\div 2
\end{array}\right.$
$\sim\left[\begin{array}{llll}
2 & 0 & 2 & -1\\
6 & 4 & 8 & 1\\
-8 & -2 & -9 & 2
\end{array}\right]\left[\begin{array}{l}
.\\
-3R_{1}.\\
+4R_{1}
\end{array}\right.$
$\sim\left[\begin{array}{llll}
2 & 0 & 2 & -1\\
0 & 4 & 2 & 4\\
0 & -2 & -1 & -2
\end{array}\right]\left[\begin{array}{l}
.\\
\div 2.\\
.
\end{array}\right.$
$\sim\left[\begin{array}{llll}
2 & 0 & 2 & -1\\
0 & 2 & 1 & 2\\
0 & -2 & -1 & -2
\end{array}\right]\left[\begin{array}{l}
.\\
.\\
+R_{2}.
\end{array}\right.$
$\sim\left[\begin{array}{llll}
2 & 0 & 2 & -1\\
0 & 2 & 1 & 2\\
0 & 0 & 0 & 0
\end{array}\right]$
... the system is consistent, , so $ w\in$Col A.
Is Aw=0?
$\left[\begin{array}{lll}
-8 & -2 & -9\\
6 & 4 & 8\\
4 & 0 & 4
\end{array}\right]\left[\begin{array}{l}
2\\
1\\
-2
\end{array}\right]$
=$\left[\begin{array}{l}
-16-2+18\\
12+4-16\\
8+0-8
\end{array}\right]=\left[\begin{array}{l}
0\\
0\\
0
\end{array}\right]$
Yes. $ w\in$Nul A.