Giving a basis for the column space of A

The columns corresponding to the pivots of your original matrix will be a basis for the column space.

The column space is the span of the column vectors of $A$, The original matrix. As got it--thanks commented, row reduction does not preserve the column space. So the column space is:

$\mathrm{span}\bigg(\begin{bmatrix} 3 \\ 3 \\ -2 \\ -2 \\ 4 \end{bmatrix}, \begin{bmatrix} 3 \\ 5 \\ 4 \\ -4 \\ 9 \end{bmatrix}, \begin{bmatrix} 3 \\ 1 \\ -8 \\ 0 \\ -1 \end{bmatrix}\bigg)$.

To find an actual basis for the column space, we need to reduce this list to a linearly independent list, if it is not already.

In fact, you can show that these three vectors are not linearly independent. Particularly, the third can be written as a linear combination of the first two.

Since the first two are linearly independent (which you should verify), we can write $\bigg(\begin{bmatrix} 3 \\ 3 \\ -2 \\ -2 \\ 4 \end{bmatrix}, \begin{bmatrix} 3 \\ 5 \\ 4 \\ -4 \\ 9 \end{bmatrix}\bigg)$ as a basis for the column space.

Fact. Let $A$ be a matrix. The nonzero rows of $\DeclareMathOperator{rref}{rref}\rref(A^\top)$ form a basis of the column space of $A$.

In our case we have $$ A= \left[\begin{array}{rrr} 3 & 3 & 3 \\ 3 & 5 & 1 \\ -2 & 4 & -8 \\ -2 & -4 & 0 \\ 4 & 9 & -1 \end{array}\right] $$ Row-reducing $A^\top$ gives $$ \rref(A^\top)= \left[\begin{array}{rrrrr} 1 & 0 & -\frac{11}{3} & \frac{1}{3} & -\frac{7}{6} \\ 0 & 1 & 3 & -1 & \frac{5}{2} \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] $$ The fact above implies that \begin{align*} \langle1, 0, -{11}/{3}, {1}/{3}, -{7}/{6} \rangle && \langle0,1, 3, -1, 5/2 \rangle \end{align*} forms a basis of the column space of $A$.