What you've determined is that $Ax=b$ has a solution if and only if $2b_1 - 2b_2 - b_3=0$. However, $Ax=b$ has a solution if and only if $b$ is in the column space of $A$.
So, an equation for the column space of $A$ is
$$
2x-2 y - z=0
$$
...or at least, this would be a way to find the equation of the plane if you had row-reduced correctly.
I think what you have claimed is true i.e., Ker $B = $ Col $A$, where $B$ is the matrix representation of the homogeneous system you mention.
Clearly, Col $A \subseteq$ Ker $B$ since if $v \notin$ Ker $B$, there will be a "zero row" in the reduced system that gives a contradiction (i.e., the LHS will be zero and RHS non-zero), so that $v \notin$ Col $A$.
Also, Ker $B \subseteq$ Col $A$, since if $v \in$ Ker $B$ the "zero rows" are now all satisfied, and the reduced system clearly has a solution (e.g., set the pivot variables equal to the RHS and set the free variables to $0$.) This solution gives $Ax = v$ so that $v \in$ Col $A$.
You may want to note that it's probably easier to use the reduced system to determine a basis for Col $A$ than it would be to try to find a basis for Ker $B$.
A more intuitive explanation
Let $A$ be $m$ x $n$. We begin with
$$\left[
\begin{array}{c|c} A & I \end{array} \right], \tag{1}$$
We then row reduce this by making the last row $0$ on the LHS. In particular, we have
$$a_m + \sum_{i=1}^{m-1} c_i a_i = 0$$
where $a_i$ denotes the $i^{th}$ row of $A$. This means
$$\begin{bmatrix} c_1 \\ \vdots \\ c_{m-1} \\ 1 \end{bmatrix} \in Null A^T$$
On the RHS, the bottom row simply becomes $$\begin{bmatrix} c_1 & \cdots & c_{m-1} & 1 \end{bmatrix} $$
Continuing in this fashion we find that row $k + 1$ on the RHS will be $$\begin{bmatrix} c'_1 & \cdots & c'_{k} & 1 & 0 & \cdots & 0 \end{bmatrix} $$
where we define $k$ to be the rank of $A$. Since these vectors are independent, we see they are a basis for Null $A^T$.
Therefore, since having a dot product of $0$ with these vectors is equivalent to being in Col $A$, we see why Col $A = $ Null $B$ (where again $B$ is the submatrix consisting of the bottom $m-k$ rows of the RHS of $(1)$ after row reduction)
Intuitively, these row reductions extract information about Col $A$ by placing vectors associated with Null $A^T$ into the RHS of $(1)$.
$\square$
Best Answer
You can prove the converse as well. If $b_3+b_2-5b_1=0$ then we will be able to solve the augmented system with infinitely many solutions and hence $b$ will belong to the column space. As an example, $A[0,0,5b_2/4-3b_1,2b_1-3b_2/4]^T=b$.