Question about intersection of planes I have the following here...

Let $\mathcal{P}_1:a_1x+b_1y+c_1z=d_1$ and $\mathcal{P}_2:a_2x+b_2y+c_2z=d_2$ be planes in $\mathbb{R}^3$. Use Linear algebra to prove that the intersection of these two planes is either empty, a line, or a plane but not a singular point. Note that since $\mathcal{P}_1$ is a plane, one of $a_1,b_1,c_1$ must be nonzero. The same is true for $\mathcal{P}_2$.

The student I am helping has only done a few weeks of linear algebra and does not know the concept of rank or dimension. All they really know is RREF, linear independence, span and finding euations of lines and planes in different forms (Vector, parametric, general etc...)
To show the intersection of two planes could be a line:
$c_1 z = d_1-a_1 x - b_1 y$ which implies $z=\frac{d_1-a_1x-b_1y}{c1}$
and $c_2 z = d_2-a_2 x - b_2 y$ which implies $z=\frac{d_2-a_2x-b_2y}{c2}$
By equating them, I get $z=\frac{d_1-a_1x-b_1y}{c1}=\frac{d_2-a_2x-b_2y}{c2}$
So if I solve for $y$, I get:
$y=\frac{(a_1c_2-a_2c_1)}{(b_2c_1-b_1c_2)}x+\frac{(c_1d_2-c_2d_1}{(b_2c_1-b_1c_2)}$
which is a line.
To show the intersection of two planes could be a Plane:
For a plane I reasoned the following. If the intersection of two planes is a plane, then $d_1 = d_2$ meaning that $a_1x+b_1y+c_1z=a_2x+b_2y+c_2z$ so that means $(a_1-a_2)x+(b_1-b_2)y+(c_1-c_2)z=0$ which is a plane.
I'm not sure if this is correct though...
What do they mean by "empty point" though?
 A: I think what you wrote for proving that the intersection could be a plane is not okay.
The intersection could not be all $\mathbb{R}^3$ because one of them are really a plane (you are supposing one of the $a_i, b_i, c_i$ is non zero).
Let us suppose that the two vectors $v_1=(a_1, b_1,  c_1 ,  d_1)$ and $v_2=(a_2, b_2,  c_2 ,  d_2)$ are multiple between them (so linearly dependent)
$v_2=\lambda v_1$ (I’m supposing that the first plane is really a plane)
Then the intersection is of course a plane, whose equation is that one of the first plane.
Let us suppose that they are not multiple between them (so that they are linearly independent). Then
$\lambda_1 v_1+\lambda_2v_2 =0 $ implies $\lambda_1=\lambda_2=0$. In this case you can write two of the variables (as you did it) in function of the third one, to getting a one dimensional space. (In fact it will be a sub space depending by only one parameter).
Hence there is no way to getting just a single point.
A: Note that the number of non-zero rows in RREF($X$) equals the rank of matrix $X$, you may find it easy to look at this problem using augmented matrix $C=[A:B]$; where $A$ is the $2\times 3$ coefficient matrix for the given system and $B$ is $2\times 1$ column vector i.e.
$C=\begin{pmatrix}a_1&b_1&c_1:d_1\\a_2&b_2&c_2:d_2\end{pmatrix}$ 
so that the following cases may arise:
(i) RREF($C$)=$\begin{pmatrix}a_1&b_1&c_1:d_1\\0&0&0:*\end{pmatrix}$ which implies rank($A$)<rank($C$) i.e. NO SOLUTION or empty set.
(ii) RREF($C$)=$\begin{pmatrix}a_1&b_1&c_1:d_1\\0&0&*:*\end{pmatrix}$ which implies rank($A$)=rank($C$)<n (n0. of variables) i.e. INFINITELY MANY SOLUTIONS; out of which exactly $n-rank(A)+1=2$ solutions are linearly independent. These $2$ LI solutions i.e. $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ can form a line in $\mathbb R^3$.
(iii)RREF($C$)=$\begin{pmatrix}a_1&b_1&c_1:d_1\\0&0&0:0\end{pmatrix}$ which implies rank($A$)=rank($C$)<n (n0. of variables) i.e. INFINITELY MANY SOLUTIONS; out of which exactly $n-rank(A)+1=3$ solutions are linearly independent. These $3$ LI solutions i.e. $(x_1,y_1,z_1)$, $(x_2,y_2,z_2)$ and $(x_2,y_2,z_2)$ can form a plane  in $\mathbb R^3$.

Note that you cannot have $rank(A)=rank(C)=n$,  so the possibility of UNIQUE SOLUTION i.e. a single point is discarded.

