In $R^3$, Given the plane $\pi : ax + by +cz + d = 0$ and the planes $\alpha : y + z = 2, \quad \beta: x - y + z = 0$ . Do there exist values of $a,b,c,d$ s.t. the three planes meet two by two in a line and the remaining three lines described by their intersection are parallel to each other?
I have tried writing the lines in parametric form to extract the vectors parallel to the lines and seeing if there are values for which all vectors are proportional but I am unsure on the results, or If I am utilizing all the information.
Any straightforward ways to do this?
Given two intersecting planes, you can always find a third plane so that your scenario is true. Think of looking at the planes "from the side" so that they look like intersecting lines. You then want a third plane so that it (from that perspective) looks like a line completing a kind of triangle.
From this viewpoint it should be obvious that, yes, there is a solution. In fact, there are infinitely many solutions!
To find a solution, take the normal vectors for your two planes: $\langle 0,1,1 \rangle$ and $\langle 1,-1,1 \rangle$ and cross them: $\langle 2,1,-1 \rangle$. This new vector is parallel to both planes.
Now again imagine looking at our three planes from the side (we can see this "triangle") the side of the triangle made by our missing plane will give us a vector parallel to the missing plane and allow us to finish finding a normal vector for our new plane. To get this side of our triangle just join a point from one of the given planes to a point on the other (make sure these points aren't on the line of intersection). Say $(0,2,0)$ on plane $\alpha$ and $(1,1,0)$ on plane $\beta$ (note: each point lies on one plane but not the other). Form a vector: $\langle 1-0,1-2,0-0 \rangle = \langle 1,-1,0 \rangle$.
Now $\langle 2,1,-1 \rangle \times \langle 1,-1,0 \rangle = \langle -1,-1,-3 \rangle$. This gives us a normal vector for our desired plane.
Finally, we just need to make sure that our plane doesn't contain the line of intersection of the first two planes. Since those planes don't (both) contain the origin, let's choose that. It looks like $-x-y-3z=0$ (i.e. $x+y+3z=0$) works.