Column space and equation of the plane

The column space of A is a plane. I need to find the equation of the plane.

Matrix $A$
$\begin{bmatrix} 2&4&6&4 \\2&5&7&6 \\2&3&5&2 \end{bmatrix}$

My solution:

my logic is that vector b which is in the Col(A)

$\begin{bmatrix} b_1 \\b_2 \\b_3 \end{bmatrix}$

Then Ax = b has a solution:

then,

$\begin{bmatrix} 2&4&6&4 = b_1 \\2&5&7&6 = b_2 \\2&3&5&2 = b_3 \end{bmatrix}$

when I reduced the matrix I produced(not sure if I did this part correctly)

$\begin{bmatrix} 1&2&3&2 = 1/2b_1 \\0&1&1&2 = b_2-b_1 \\0&0&0&0 = 2b_1-2b_2-b_3 \end{bmatrix}$

Not quite sure where to go from here or if my workings are correct to this point. Looking for some guidance

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.