Distance of a plane `2x+y+2z+5=0` from a point `(2,1,0)` is

find distance of the plane 2x-y+3z=4 from point (1,0,-2)

Distance of the plane 2x-y+2z+3=0 from (3,2,-1) is

Distance of the Plane 2x-y+2z+3=0 from (3,2,-1) is

A plane passes through (1,-2,1) and is perpendicualr to two planes 2x-2y+z=0" and "x-y+2z=4, then the distance of the plane from the point (1,2,2) is

A plane passes through (1, -2, 1) and is perpendicular to two planes 2x-2y+z=0 and x-y+2z=4. The distance of the plane from the point (0, 2, 2) is

Find the equation of the plane through the line of intersection of the planes x+y+z=1 and 2x+3y+4z=5 which is perpendicular to the plane x-y+z=0 Then find the distance of plane thus obtained from the point A(1,3,6) .

The square of the distance of the point of intersection of the line and the plane (x-1)/(2)=(y-2)/(3)=(z+1)/(6) and the plane 2x-y+z=6 from the point (- 1, -1, 2) is_________