I have some difficulties with a problem, where i have been given four planes: $$x+ay+az=a$$ $$x+a^2y=a^3$$ $$x+a^2y+az=a$$ $$x+ay+2az=a,$$ and need to find the points in $M$, where any point belongs to $M$, if it contains the coordinates $4$ and $-4$, and there is an $a$ in which it lies in all four planes.
I know how to reduce the system of equations to reduced row echelon form, and have also found values of $a$, where they all intersect, but i can't seem to find a way to solve this specific problem. Any help would be appreciated.
If I understood your question correctly, I believe you are looking for any points satisfying the following conditions:
(i) There must be at least one $\:4\:$ and one $\:-4\:$ among the coordinates; and
(ii) There must be at least one $\:a\:$ such that the point lies on all four planes.
We start by simplifying the conditions.
a) If $\:a=0\:$, then all equations become $\:x=0\:$, which yields the four points $\:(0,\pm 4, \pm 4)\:$.
b) If $\:a \neq 0\:$, then combining the first and last equations gives $\:z=0\:$, and combining the second and third equations gives $\:a=\pm 1\:$. If $\:a=1\:$, all four equations become $\:x+y=1\:$; if $\:a=-1\:$, we get the system $\:x-y=-1\:$ and $\:x+y=-1\:$. However, neither case allows for a solution containing a $\:4\:$ and a $\:-4\:$.
So under the stated conditions, the only solutions would be the four points $\:(0,\pm 4, \pm 4)\:$.
However, if the conditions were loosened slightly so that the first condition became
(i) There must be at least one $\:4\:$ or one $\:-4\:$ among the coordinates
then case a) would yield all the points on the four lines $\:x=0, y=\pm 4\:$ and $\:x=0, z=\pm 4\:$;
and case b) with $\:a=1\:$ would yield the points: $\:(5,-4,0),(4,-3,0),(-3,4,0)\:$ and $\:(-4,5,0)\:$.