Find the equations of the planes parallel to the plane x – 2y + 2z – 3 = 0 and which are at a unit distance from the point (1, 1, 1).
Since the planes are parallel to x – 2y + 2z – 3 = 0, they must be of the form:
x – 2y + 2z + θ = 0
We know, the distance of point (x1,y1,z1) from the plane
is given by:
According to the question, the distance of the planes from (1, 1, 1) is 1 unit.
⟹
⟹
⟹ or
⟹ θ = 2 or –4
⟹ The required planes are:
x – 2y + 2z + 2 = 0 and x – 2y + 2z – 4 = 0