This question is quite bothering me so I want to correct myself:
What I think of the column space as the span of individual dimention, so in the following X1 and X2 are Individual dimensions as follows: $$\begin{pmatrix} 1 & 2 & 5\\ 6 & 5 & 3\\ 3 & 4 & 2 \end{pmatrix}\begin{pmatrix} X_1 \\ X_2 \\ X_3 \\ \end{pmatrix}$$
So the question is if we perform RREF of some arbitrary matrix and get column 1 and column 2 as the pivots. $$\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{pmatrix}$$ Does it mean that $X_3$ useless because it is not the basis as well as the third column is useless?
If you had made up a valid question, however, the answer is that if you can write $\vec b$ as a linear combination of the columns of $A$, i.e., you can solve $A\vec x=\vec b$, then — yes — you could write $\vec b$ as a linear combination of just the pivot columns. That is, we can rewrite the non-pivot columns as linear combinations of the pivot columns and therefore express $\vec b$ as a linear combination of just the pivot columns. This means that there will be a solution $X$ with each of the free (non-pivot) variables equal to $0$.
Geometrically, if you have a $2$-dimensional column space (as is the case with your RREF), then for any $\vec b$ in that column space, we get a line of solutions. What I said earlier is that this line must intersect the plane $x_3=0$. The line of solutions is parallel to the nullspace of the matrix. This nullspace cannot be contained in the plane $x_3=0$ unless one of the pivots is in the third column. (By the way, your RREF should in general have nonzero entries in the first two positions of the third column. Better to work with the general case here.)