Find the equations of the planes parallel to the plane `x - 2y + 2z -4 = 0`, which are at a unit distance from the point (1, 2, 3).

Options

A. `x + 2y + 2z = - 6, x + 2y + 2z = 5 `
B. ` x -2y - 6 = 0, x - 2y + z = 6 `
C. ` x + 2y + 2z = 6, x + 2y + 2z = 0 `
D. `x - 2y+ 2z = 0, x - 2y + 2z - 6 =0`

Answer

D

Explanation

Given equations of plans
`x - 2y + 2z + 4=0" "……..(i)`
Equations of plane parallel to plane (i) is
`x -2y + 2z + k = 0 " "…..(ii) `
Since, distance of plane (ii) from point (1,2,3) is 1 unit
`therefore " " (|1(1) -2(2) +2(3) +k|)/(sqrt((1)^(2) +(-2)^(2) +(2)^(2)))=1`
`rArr " "(|1-4 + 6 +k|)/(sqrt(1+4+4)) =1`
`rArr " "(|3+k|)/(sqrt(9))=1`
`rArr " "|3+k| = 3 rArr 3 + k = pm 3`
`rArr" "k = 0,-6`
`thereofre ` Required equation of plane are
`x - 2y + 2 +0=0 and x -2y + 2z - 6 =0 `
`rArr x - 2y + 2z = 0 and x -2y + 2z - 6 =0`