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?