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


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