Intersection of 2D planes in 4D space

Let $\mathbf{n}, \mathbf{m}$ be two linearly independent vectors in $\mathbb{R}^4$. A plane $P$ is given by $$\mathbf{n}\cdot (\mathbf{r}-\mathbf{r}_0)= \mathbf{m}\cdot (\mathbf{r}-\mathbf{r}_0)=0 $$ and the second plane $P'$ is given by $$\mathbf{n}'\cdot (\mathbf{r}-\mathbf{r}_0')= \mathbf{m}'\cdot (\mathbf{r}-\mathbf{r}_0')=0 $$ note that $\mathbf{r}_0$ and $\mathbf{r}'_0$ can be changed to any other point on the plane and the equations still describe the same plane. Hence if the intersection is non-empty we can take $\mathbf{r}_0=\mathbf{r}_0'$. Then we only need to find the solutions to $$ A\cdot (\mathbf{r}-\mathbf{r}_0):= \begin{pmatrix} \mathbf{n}\\ \mathbf{m}\\ \mathbf{n}'\\ \mathbf{m}' \end{pmatrix} \cdot (\mathbf{r}-\mathbf{r}_0)=0 $$ Note that without loss of generality we can choose $\mathbf{n}', \mathbf{m}'$ to either not lie in $V=\mathrm{span}(\mathbf{n},\mathbf{m})$, or only one of them lies there or both.

Now if $\mathbf{n}', \mathbf{m}'$ do not lie in the subspace spanned by $\mathbf{n},\mathbf{m}$ then this matrix is invertible, meaning the intersection is $\mathbf{r}_0$ (a point).

If one of $\mathbf{n}',\mathbf{m}'$ is in the span of $\mathbf{n},\mathbf{m}$ then the kernel of this linear transformation is 1-dimensional and the intersection becomes line.

If both $\mathbf{n}', \mathbf{m}'$ lie in $V$ since they also share a common point $\mathbf{r}_0$ by assumption then the two planes coincide. You should exclude this possibility if your original planes are distinct.

So finally we need to understand when the intersection is empty (we assumed it is non-empty above). This only happens if the planes are parallel. If they are not parallel then the rank of $A$ is at least 3. You can investigate that $$ A\mathbf{r} = \begin{pmatrix}\mathbf{n}\cdot \mathbf{r}_0\\ \mathbf{m}\cdot \mathbf{r}_0 \\ \mathbf{n}'\cdot \mathbf{r}'_0\\ \mathbf{m}'\cdot \mathbf{r}'_0\end{pmatrix} $$ has a solution if rank $A$ is at least 3. So we must have rank $A$ at most 2. But then $A$ comes from the equations of $P,P'$ so it has rank at least two. This means rank $A$ is necessarily $2$, and hence the planes are parallel.