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

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

Find the distance of the plane 2x-y-2z-9=0 from the origin.

Find the distance of the plane 2x - 3y + 4z - 6 = 0 from the origin.

Find the distance of the plane 2x 3y + 4z 6 = 0 from the origin.

If d_(1),d_(2),d_(3) denote the distances of the plane 2x-3y+4z+2=0 from the planes 2x-3y+4z+6=0 ,4x-6y+8z+3=0 and 2x-3y+4z-6=0 respectively, then

The distance of the plane x+2y-z=2 from the point (2, -1, 3) , measured in the direction with the direction ratios (2, 2,1) is

If d_(1),d_(2),d_(3) denote the distances of the plane 2x-3y+4z=0 from the planes 2x-3y+4z+6=0 4x-6y+8z+3=0 and 2x-3y+4z-6=0 respectively, then

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 P = 0 passing through the point (1, 1, 1) is perpendicular to the planes 2x-y+2z=5 and 3x+6y-2z=7 . If the distance of the point (1, 2, 3) from the plane P = 0 is k units, then the value of 34k^(2) is equal to

Show that the distance between planes 2x-2y+z+3=0 and 4x-4y+2z+5=0 is 1/6