Finding more than one basis for a column space Let's say the matrix A is defined as the following:
$$A=\begin{pmatrix}6 & 3 & -1 & 0 \\ 1 & 1 & 0 & 4 \\ -2 & 5 & 0 & 2\end{pmatrix}$$
Whose RRE form is:
$$A_{RRE}=\begin{pmatrix}1 & 0 & 0 & * \\ 0 & 1 & 0 & * \\ 0 & 0 & 1 & *\end{pmatrix}$$
Then I can write a basis for the column space of matrix A using the three linearly independent vectors as such:
$$B_1=\left\{\begin{pmatrix}6\\ 1 \\ -2\end{pmatrix},\begin{pmatrix}3\\ 1 \\ 5\end{pmatrix},\begin{pmatrix}-1\\ 0 \\ 0\end{pmatrix}\right\}$$
However, what if I want to find a different basis for colA besides the one above, such that no vector in the basis is a scalar multiple of any of the vectors in B1? Is this method valid?
1. Re-arrange columns of matrix, which won't change the column span:
$$A'=\begin{pmatrix}0 & 6 & 3 & -1 \\ 4 & 1 & 1 & 0 \\ 2 & -2 & 5 & 0\end{pmatrix}$$
2. Put into RRE form:
$$A'_{RRE}=\begin{pmatrix}1 & 0 & 0 & * \\ 0 & 1 & 0 & * \\ 0 & 0 & 1 & *\end{pmatrix}$$
3. Preliminary basis:
$$B'_2=\left\{\begin{pmatrix}0\\ 4 \\ 2\end{pmatrix},\begin{pmatrix}6\\ 1 \\ -2\end{pmatrix},\begin{pmatrix}3\\ 1 \\ 5\end{pmatrix}\right\}$$
4. Take linear combinations of above basis:
$$B_2=\left\{\begin{pmatrix}0\\ 4 \\ 2\end{pmatrix},\begin{pmatrix}6\\ 5 \\ 0\end{pmatrix},\begin{pmatrix}9\\ 2 \\ 3\end{pmatrix}\right\}$$
I know a better method would be to just take linear combinations of the existing basis B1, but I want to know if this method works and if not, why.

Yes it works fine and all the basis you have found are correct as any triple of linearly independent vectors. 
A simpler choice would be: $v_1=(1,1,0)$, $v_2=(0,1,0)$, $v_3=(0,0,1)$.
