Find the vector equation of the plane which is at a distance of 5 units from the origin and which is normal to the vector `2hati + hatj + 2hatk`.

Explanation

Let `barn = 2hati + hatj + 2hatk` be a normal vector to the plane.
Unit vector along the normal to the plane is given by
`hatn=(barn)/(|barn|)=(2hati+hatj+2hatk)/(sqrt(2^(2)+1^(2)+2^(2)))=(2hati+hatj+2hatk)/(3)`
Equation of the plane having unit vector `hatn` along the normal to the plane and at a distance pf p units from the origin is given by `barr*hatn = p` ,
`therefore` equation of the required plane is `barr*((2hati+hatj+2hatk)/(3))=5`
i.e., `barr*(2hati + htaj + 2hatk) =15` .