Answer :

Formula : Plane = r . (n) = d


Where r = any random point


n = normal vector of plane


d = distance of plane from origin


If two planes are parallel , then their normal vectors are same


Therefore ,


Parallel Plane x – 2y + 2z – 3 = 0


Normal vector = (i - 2j + 2k)


Normal vector of required plane = (i - 2j + 2k)


Equation of required planes r . (i - 2j + 2k) = d


In cartesian form x – 2y + 2y = d


It should be at unit distance from point (1,1,1)


Distance






For + sign = > 3 = 1 - d = > d = - 2


For - sign = > 3 = - 1 + d = > d = 4


Therefore equations of planes are : -


For d = - 2 For d = 4


x – 2y + 2y = d x – 2y + 2y = d


x – 2y + 2y = - 2 x – 2y + 2y = 4


x – 2y + 2y + 2 = 0 x – 2y + 2y – 4 = 0


Required planes = x – 2y + 2y + 2 = 0


x – 2y + 2y – 4 = 0


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