Assuming that the column vectors of your matrix $A$ can be written with $v=\begin{bmatrix}3&-2&1\end{bmatrix}^T$ and $w=\begin{bmatrix}-5&6&1\end{bmatrix}^T$ where $A=[v\quad w]$ then your (linear independant) vectors $v$ and $w$ span obviously one plane $P=\operatorname{span}\{v,w\}\subset\mathbb{R}^3$ with $\dim(P) = 2$.
Assume that there is one vector $x\in P$. $x$ has then the following form $x=\lambda v+\mu w\in P$ and therefore can be written as a linear combination of the two vectors $v,w$ that span $P$. If you want to check whether $u=\begin{bmatrix}0&4&4\end{bmatrix}^T$ is in $P$ you have to find $\lambda,\mu$ such that the combination of the vectors $v,w$ equals $u$. If there is no solution, then the vector is not part of the plane.
Basically you solve the following system of equations:
$\begin{bmatrix}3&-5\\-2&6\\1&1\end{bmatrix}\cdot\begin{bmatrix}\lambda\\\mu\end{bmatrix}=\begin{bmatrix}0\\4\\4\end{bmatrix}$
Hint: This system has a unique solution.