Find a plane in $\mathbb{R}^4$ skewed to another 2 planes Given $\pi_1,\pi_2$ planes in $\mathbb{R}^4$ as follows:
\begin{align*}
 \pi_1 &= <(1,0,-4,-3), (0,1,-1,-2)> + (0,0,1,0) \\
 \pi_2 &= <(2,0,-3,-3),(1,0,1,0)> + (1,0,0,0)
\end{align*}
Find another plane $\pi \subset \mathbb{R}^4$ skewed to $\pi_1, \pi_2$ if exists or prove it doesn't otherwise. 

My attempt: If $\pi = <v1, v2> + p$, i tried taking $v1,v2$ from $<(0,0,1,0) - (1,0,0,0)> ^ \perp$ not both contained in the subspace asociated to $\pi_1$ and $\pi_2$ and $p = \frac{(0,0,1,0) + (1,0,0,0)}{2}$ but it didn't worked since it had intersection with $\pi_2$. So, i run out of ideas here. Any ideas?
 A: Let $\Lambda_1$ and $\Lambda_2$ be two affine planes in $\Bbb R^4$. They are parallel to linear subspaces $\Pi_1$ and $\Pi_2$ and we may write them as cosets $\Lambda_1=a_1+\Pi_1$ and $\Lambda_2=a_2+\Pi_2$.
If the affine planes are not parallel, then $\Pi_1\ne\Pi_2$ so that $\Pi_1+\Pi_2$ must be either a hyperplane or the whole space $\mathbb{R}^4$. In the former case, that means the two linear subspaces intersect nontrivially at a line $\ell$, so that the two affine planes contain a pair of parallel lines $a_1+\ell$ and $a_2+\ell$. This may seem contrary to how we want to define "skew," but the linear subspaces' spanning a proper subspace turns out to be a necessary condition.
Proposition. $\Lambda_1$ and $\Lambda_2$ don't intersect $\iff a_1-a_2\not\in\Pi_1+\Pi_2$.
A point of intersection $a_1+v_1=a_2+v_2$ would indicate $a_1-a_2\in\Pi_1+\Pi_2$ and conversely.
If $\Pi_3$ is any plane in $\Pi_1+\Pi_2$ which is not the same as $\Pi_1$ and $\Pi_2$ then $\frac{1}{2}(a_1+a_2)+\Pi_3$ will end up being a plane that is skew to the first two planes, which should be doable in this case.
The "most symmetric" answer should be "halfway" between the two original planes, which should come from picking $\Pi_3$ "halfway" between $\Pi_1$ and $\Pi_2$ within their 3D span.
