Finding the plane that contains all the lines of a given form This is a bit from a previous questions that didn't get any attention:
in the affine space $\Bbb A^4$ consider the lines $r_k:\begin{cases}x=1+t(2k-1) \\ y=2kt \\ w=3+3t(2k-1) \\ z=1+t(2k-1)\end{cases} (t\in \Bbb R).$
The exercise asks me to find the reciprocal position between the plane $\omega$ that contains all the $r_k$ and the plane $\pi:\begin{cases}x-y+w-1=0 \\ x-2y+z+1=0\end{cases}$ 
Here's my reasonning: two or more lines lie in the same plane if and only if they intersect (or coincide), So $\omega$ is the plane passing through $r_0\begin{cases}x=1-t \\ y=0 \\ w=3(1-t) \\ z=1-t\end{cases} (t\in \Bbb R)$ and $C=r_0 \cap r_1=(1,0,3,1).$ Now, exactly one line passes through two points, so once we choose any $A, B\in r_0$ different from $R$, $\omega$ is the plane that passes through $A,B,R$. We pick $A=(0,0,0,0), B=(-1,0,-3,-1), C=(1,0,3,1).$ 

It feels like $\omega$ should be a sort of "diagonal" of $\Bbb A^4$, but how do I express it without resorting to the normal as I do below?

We have $AB=B-A=(-1,0,-3,1), AC=(1,0,3,1)$. In order to find the normal $N=(a,b,c,d)$ we put $$\begin{cases}(a,b,c,d)\cdot(-1,0,-3,1)=0 \\ (a,b,c,d)\cdot(1,0,3,1)=0\end{cases} \iff a+3c+d=0.$$
So now, I've been looking at similar questions here and it seems I'm supposed to choose arbitrary (non-zero?) values that satisfy the above? In this case I'd take $a=b=1, c=-1, d=2$. And since $\omega$ passes through the origin, its equation should be $x+y-w+2z=0$. Then, if I put this and the equations of $\pi$ in a system, I get the solution $\begin{cases}x=t\in\Bbb R \\ y=\frac34 \\ w=-t+\frac74 \\ z=-t+\frac12 \end{cases}$... however this must be wrong because it doesn't contain any of $A,B,C$...
 A: You started off OK. All of the lines obviously pass through $(1,0,3,1)$, so that’s one known point on the plane. You need another two points to determine a plane uniquely, but those three points must not be colinear. That’s not the case for the three points you’ve chosen. You need to choose the third point from a line other than the one through $A$ and $C$. Since the plane passes through the origin, you can use the direction vectors of any two distinct lines in the family as the other two points. Setting $k=1$ and $t=1$ for instance produces $(2,2,6,2)$, which will do.  
You can produce a parameterization of the common plane directly from these three points: $\lambda (1,0,3,1) + \mu (2,2,6,2)$. Representing this plane via implicit Cartesian equations will require a system of two linear equations—a plane in $\mathbb A^4$ is the intersection of two hyperplanes. The plane passes through the origin, so the equations are homogeneous, of the form $\mathbf n^T\mathbf x=0$. These equations say that the plane can be characterized by a linearly-independent pair of vectors orthogonal to it, i.e., a basis for the plane’s orthogonal complement. This is also the null space of $$\begin{bmatrix}1&0&3&1\\2&2&6&2\end{bmatrix},$$ which can be found using any of the usual methods. For instance, the rref of this matrix is $$\begin{bmatrix}1&0&3&1\\0&1&0&0\end{bmatrix},$$ from which we can read the basis vectors $(3,0,-1,0)$ and $(1,0,0,-1)$. These correspond to the equations $3x-w=0$ and $x-z=0$. You can verify by inspection that the original line equations satisfy these two.  
If you hadn’t happened to notice that the plane passes through the origin, you could proceed in a similar fashion by passing to homogeneous coordinates. Three noncolinear points are still needed to determine the plane, but now you can use the known intersection point $(1,0,3,1)$ along with two points at infinity—the homogeneous counterparts of the lines’ direction vectors. Taking, say, the points at infinity for $k=0$ and $k=1$, compute the null space of $$\begin{bmatrix}1&0&3&1&1\\-1&0&-3&-1&0\\1&2&3&1&0\end{bmatrix}.$$ From the rref of this matrix we can find the basis vectors $(1,0,0,-1,0)$ and $(3,0,-1,0,0)$, which match the two implicit Cartesian equations found previously.
