Find the equations of the plane parallel to the plane `x + 2y + 2z + 8 = 0`
which are at a distance of 2 units from the point (1, 1, 2).

Explanation

The equation of the plane parallel to the plane `x + 2y + 2z + 8 = 0` is
`x + 2y + 2z + lambda = 0" "…(1)`
Nows, distance of this plane from the point (1, 1, 2)
`|(1(1)+2(1)+2(2)=lambda)/(sqrt(1^(2)+2^(2)+2^(2)))|`
`=|(1+2+4+lambda)/(sqrt(1+4+4))|=|(lamda+7)/(3)|`
But this distance is given to be 2 units.
`therefore | (lambda + 7)/(3)| =2`
` therefore (lambda + 7)/(3) = pm 2`
`therefore (lambda+7)/(3)=2 or (lambda+7)/(3)=-2`
`therefore lambda+7=6 or lambda+7=-6`
`therefore lambda=-1 or lambda=-13`
Hence, the equation of the required planes are
`x + 2y + 2z -1 = 0 and x + 2y + 2z - 13 = 0`.