Is my proof complete? I feel like something is missing.
What I have written:
Consider $\vec{b} \in \mathbb{R}^3$, without loss of generality.
Then $[A|\vec{b}] = \begin{bmatrix} a_1 & ... & a_{1n}|&b_1 \\ a_2 & ... & a_{2n}|&b_2 \\ a_3 & ... & a_{3n}|&b_3 \end{bmatrix}$
In order for the solution set to be a plane through the origin, $b_1 = b_2 = b_3 = 0$. However, $\vec{b}\neq\vec{0}$, so then $[A|\vec{b}]$ cannot be a plane through the origin.