Since the plane is parallel to the x-axis, a radius from the origin or the z-axis out to the plane will be perpendicular to the plane. So, a random point (x,y,z) can be on the plane at the end of the origin radius.
If you construct 2 vectors in the plane, one connected to the point (x,y,z) to one of the points and then a vector from given point to point, you have 2 vectors that, when subjected to the cross product, you get a new vector perpendicular to the plane.
That vector can then be "dotted" with the x-axis to produce the equation of the plane.
If P1= (3,1,1) & P2= )1,-2,3), vector P1P2= (-2,-3,2) ,then the vector from P2 to (x,y,z) = {(x-1),(y+2),(z-3)}
Use the cross product P1P2xP2XYZ = (-{3(z-3)+2(y+2)},{2(z-3)+2(x-1)},{3(x-1)-2(y+2)}) = V (vector)
Now i dot V yields -3(z-3) -2(y+2) = 0 Simplifying, we get z = (5/3) -(2/3)y
