Distance of the plane `4x+5y+3z = 6` from the point `(-2,-1,1)` is `(3^a+b)/(25sqrt(2))`, then (b-a) is.

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

Statement-1 The distance between the planes 4x-5y+3z=5 and 4x-5y+3z+2=0 is (3)/(5sqrt(2)) . Statement-2 The distance between ax+by+cz+d_1=0 and ax+by+cz+d_2=0 is |(d_1-d_2)/(sqrt(a^2+b^2+c^2)|.

The plane ax+by+cz=105 contain the lines 2x-y+z=3and3x+y+z=5 . If point (2 ,1, -1) is at a distance of (1)/(sqrt(6)) from the plane , then a+b+c=

The equation of a plane passing through the line of intersection of the planes x+2y+3z=2 and x-y+z=3 and at a distance (2)/(sqrt(3)) from the point (3,1,-1) is

What is the distance of the point (2,3,4) from the plane 3x-6y+2z+11=?

Show that the distance of the point of intersection of the line (x-2)/3=(y+1)/4=(z-2)/12 and the plane (x-y+z=5) from the point (-1,-5,-10) is 13 units.

The distance of the point (-2,4,-5) from the line (x+3)/(3)=(y-4)/(5)=(z+8)/(6) is

If the distance of the point (1, 2, 3) from the plane 2x - y + z = 6 measured parallel to line x/1=y/2=z/3 is sqrt(a) sqrt(b) where a and b are two prime numbers then find max. (a, b).

What is the distance of the point (2,3,4) from the plane 3x-6y + 2z + 11 = 0