With the parameter $h\in\Bbb R$, consider in the affine space $\Bbb A^4$ the system $$\begin{cases}x+2y+2z+w=5 \\x+y-hz=6 \\ -x+hy+2z+(h+1)w=h+1. \end{cases}$$Calling $\Sigma_h$ the set of solutions, I'm asked to find a hyperplane containing $\Sigma_1$ and $\Sigma_{-4}$.
Now, I've found those subspaces, but I'm not sure about the hyperplane.
The vectors of $\Sigma_1$ are of the form $(x,y,z,w)=(-1,5,-2,0)+\alpha(1,-1,0,1)$ and those of $\Sigma_{-4}$ are of the form $(x,y,z,w)=(7,-1,0,0)+\alpha(-14,6,1,0)+\beta(1,-1,0,1).$
Calling $G_1$ and $G_{-4}$ the generating sets of $\Sigma_1$ and $\Sigma_{-4}$, I noticed that $G_1\subset G_{-4}$. I thought this could mean that "starting" from $(-1,5,-2,0)$, adding to $G_{-4}$ the vector connecting $(-1,5,-2,0)$ to $(7,-1,0,0)$, i.e. $(8,-6,2,0)$, I would get the hyperplane.
I got $\{(x,y,z,w)=(-1,5,-2,0)+\alpha(-14,6,1,0)+\beta(1,-1,0,1)+\gamma(4,-3,1,0) | \alpha,\beta,\gamma\in\Bbb R\}$
Does it make sense?