In the formula of the distance between a point and a plane (not any plane but a plane parallel to the original one, equation [1] ), keeping D=2, and d as d itself, we get:
D=|ax1+by1+cz1+d|√a2+b2+c2
2=∣∣1⋅0+2⋅0+(−2)⋅(−12)+d∣∣√1+4+4
|d+1|=2⋅3 => |d+1|=6First solution:
d+1=6 => d=5
→x+2y−2z+5=0Second solution:
d+1=−6 => d=−7
→x+2y−2z−7=0
