What are the direction cosines of the normal to the plane `4x+12y+3z=65?` Also find the lenth of perpendicular from the origin to the plane.

The direction cosines of the normal to the plane x + 2y - 3z + 4 = 0 are

Find the length of the perpendicular drawn from the origin to the plane 2x \ 3y+6z+21=0.

Find the length of the perpendicular drawn from the origin to the plane 2x - 3y+6z+21=0.

Find the normal form of the plane 2x+3y-z = 5 . Also find the length of perpendicular from origin and d.c's of the normal to the plane.

Find the d.c.'s of the normal and length of perpendicular from origin to the plane x=2 .

Reduce the equation of the plane x-2y-2z=12 to normal form and hence find the length of the perpendicular for the origin to the plane. Also, find the direction cosines of the normal to the plane.

Reduce the equation 2x-3y-6z=14 to the normal form and hence fine the length of perpendicular from the origin to the plane. Also, find the direction cosines of the normal to the plane.

Find the normal form of the plane x+2y-2z+6=0 . Also find the length of perpendicular from origin to this plane and the d.c.'s of the normal.

Find the length of the perpendicular from origin to the plane vecr.(3i-4j-12k)+39=0 .

Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x-3y+4z-6=0.