Wow . . . what a stupid blunder!
Sorry, everyone . . .
As I assume Soroban tried to attempt this problem algebraically. I've choosen a completely different way to do this question:
1. From the equation of the plane you know that the normal vector is . That means is perpendicular to the x-y-plane.
2. I looked for a line which lies simultaneously in and the x-y-plane and took the line which connects the intersection points of and the x-axis and and the y-axis. The intersection points are A(-4,0,0) and B(0,-2,0).
3. I took the line as a kind of axis. I then "tilted" the x-y-plane (which is perpendicular to !) about this "axis" by 30° to get . If I tilt the x-y-plane by 30° then the normal vector of and the z-axis include an angle of 30°.
4. I took the vector and rotated it by 30° to get the vector (By the way: You probably have noticed that there must exist a second vector if you rotate the x-y-plane in the opposite direction)
5. Now I know 2 vectors which span the plane . Since has to contain the origin the parametric equation of is:
: ....... or .......
6. The cross product of the 2 direction vectors which span is the normal vector of :
And therefore the scalar equation of is:
7. I finally used derive 6.0 to draw the 2 planes and the normal vector. This program is able to rotate the 3-D-graphs about an axis. I took 2 screenshots to show how the planes and the normal are placed.