If you understand what it means in geometric terms, then you are most of the way there. I'll try to elaborate a little on the algebra involved.
The equation can be rewritten as $xv_1+yv_2+zv_3=b$, where $v_1$, $v_2$, and $v_3$ are the column vectors. The question is, based on the relationship among the columns, which vectors $b$ can you plug in to make the equation have a solution. Since there is a linear relationship among the columns, the answer, as you said, is that not all choices of $b$ will work. You have $v_2=2v_1$, so the equation can also be written as $(x+2y)v_1+zv_3=b$. Thus, $b$ must lie in the $2$-dimensional subspace spanned by $v_1$ and $v_3$, your first and third columns.
More generally, if there is some nontrivial linear relationship among the columns of the form $a_1v_1+a_2v_2+a_3v_3=0$, where $a_1$, $a_2$, and $a_3$ are scalars and at least one is nonzero, then you can write one of your vectors as a linear combination of the other $2$, and hence any vector in the span (possible "right-hand sides" in your question) will have to be in the span of these $2$. The same idea generalizes to all (finite) dimensions. If the set of column vectors is not linearly independent, then the span of the columns has lower dimension.
If you want to see explicitly what choices of $b$ will not work in your example, you can write out what the span of $v_1$ and $v_3$ looks like, and choose any vectors not in this span. For a quick example in this case, you could choose the cross product $v_1\times v_3$, since it is orthogonal to both $v_1$ and $v_3$.
Another way to look at this is that if the columns are not linearly independent, then neither are the rows, and a linear relationship among the row vectors translates to a linear relationship among the entries of the right hand side. That is, if $A$ has row vectors $r_1$, $r_2$, and $r_3$, then
$A\mathbf{x}=
\begin{bmatrix}
r_1\cdot\mathbf{x}\\
r_2\cdot\mathbf{x}\\
r_3\cdot\mathbf{x}
\end{bmatrix}
$, and if there are scalars $a_1$, $a_2$, and $a_3$, not all zero, such that $a_1r_1+a_2r_2+a_3r_3=0$, then it follows from the linearity of the dot product in each variable that $a_1(r_1\cdot\mathbf{x})+a_2(r_2\cdot\mathbf{x})+a_3(r_3\cdot\mathbf{x})=0$. That is, the right-hand side must lie in the plane described by the equation $a_1x+a_2y+a_3z=0$.