The distasnce of the plane `2x-3y+6z+14=0` from the origin is (A) 2 (B) 4 (C) 7 (D) 11

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

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

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

Find the distance of the plane 2x + 5y + sqrt7z + 12= 0 from the origin.

If theta is the angel between the planes 2x-y+z-1=0 and x-2y+z+2=0 then costheta= (A) 2/3 (B) 3/4 (C) 4/5 (D) 5/6

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

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 direction cosines of a normal to the plane 2x-3y-6z+14=0 are (A) (2/7,(-3)/7,(-6)/7) (B) ((-2)/7,3/7,6/7) (C) ((-2)/7,(-3)/3,(-6)/7) (D) none of these

The equation of the plane passing through the intersection of the planes x+2y+3z+4=0a dn 4x+3y+2z+1=0 and the origin ils (A) 3x+2y+z+1=0 (B) 3x+2y+z=0 (C) 2x+3y+z=0 (D) x+y+z=0

In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.(a) z = 2 (b) x + y + z = 1 (c) 2x + 3y z = 5 (d) 5y + 8 = 0