The vector equation of the plane which is perpendicular to `2hati-3hatj+hatk=0` and at a distance of 5units from the origin

The equation of a plane which is perpendicular to (2hati-3hatj+hatk) and at a distance of 5 units from the origin is

The equation of the plane through the intersection of the planes vecr.(2hati+6hatj)+12=0 and vecr.(3hati-hatj+4hatk)=0 and at a unit distance from the origin, is

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 .

The vector equation of the plane which is at a distance of 8 units from the origin which is normal to the vector 2hati+hatj+2hatk is

The vector form of the equation of the plane which is at a distance of 3 units from the origin and has hati + hatj - 3hatk as a normal vector, is

Find the vector equation of the plane which is at a distance of 6 units from the origin and which is normal to the vector 2hati - hatj + 2hatk .

The vector equation of a plane is vecr.(2hati+2hatj-hatk)=21 , find the length of the perpendicular from the origin to the plane.

Find the vector equation of a plane which is at a distance of 7 units from the origin and which is normal to the vector 3 hati + 5 hatj - 6 hatk . (ii) Find the vector equation of a plane , which is at a distance of 5 units from the origin and its normal vector is 2 hati - 3 hatj + 6 hatk .

Find the vector and cartesian form of the equation of the plane which is at distance of 6 units from the origin and perpendicular to 2hati+3hatj-6hatk .

Find the vector equation of a plane, which is at a distance of 5 units from the origin and its normal vector is 2hati-3hatj+6hatk .