Answer
Yes because $Ax=u$ is consistent.
Work Step by Step
If $u$ is in the plane spanned by the columns of $A$, then there exists an $x$ such that $Ax=u$. This can be determined by solving the equation through row reduction.
This is the augmented matrix of the system:
$$
\begin{bmatrix}
3&-5&0\\
-2&6&4\\
1&1&4
\end{bmatrix}
$$
For convenience, switch the third and first rows:
$$
\begin{bmatrix}
1&1&4\\
3&-5&0\\
-2&6&4\\
\end{bmatrix}
$$
Now add multiples of the first row to the second and third rows to make a pivot entry for row 1:
$$
\begin{bmatrix}
1&1&4\\
0&-8&-12\\
0&8&12\\
\end{bmatrix}
$$
Add the second row to the third:
$$
\begin{bmatrix}
1&1&4\\
0&-8&-12\\
0&0&0\\
\end{bmatrix}
$$
And divide the second row by 4:
$$
\begin{bmatrix}
1&1&4\\
0&1&1.5\\
0&0&0\\
\end{bmatrix}
$$
The matrix is in row echelon form and there is no row that reduces to $0=1$. Thus, the equation is consistent, and $u$ is in the span of the columns of $A$.
