I tried to solve this problem in the following way, I just need you to tell me if it's right.
Find a matrix $A$ such that $u$ is in $\operatorname{Null}(A)$. $$u = \begin{bmatrix} 2 \\ 1 \\ 2 \end{bmatrix}$$
I found $A$ in the following way:
$$A=\begin{bmatrix}
a&\frac{1}{2}u&u
\end{bmatrix}$$
where $a = \begin{bmatrix}
-2 \\ -3 \\ 3
\end{bmatrix}$
$$A = \begin{bmatrix}
-2&1&2 \\ -3&1/2&1 \\ 3&1&2
\end{bmatrix}$$
If I reduce it to reduced row echelon form, $\dim(\operatorname{Null}(A))=1$. Is this the correct solution?