Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the origin, then 
Distance of the point (x1,y1,z1) from the plane Ax + By + Cz = D is
⇒ 
The equation of a plane having intercepts a, b, c on the x-, y-, z- axis respectively is
⇒ 
Comparing with Ax + By + Cz = D, we get -
A = 1/a, B = 1/b, C = 1/c, D = 1
Given, the plane is at a distance of 'p' units from the origin.
So, The point is O(0,0,0)
∴ x1 = 0, y1 = 0, z1 = 0
Now,
Distance 
Substituting all values, we get -




⇒ 
squaring both sides, we get -
⇒ 
Hence Proved.