Find the distance of the plane `3x- 4y+12 z=3` from the origin.

Explanation

Distance of a plane from origin can be given as,

`D = |d/sqrt(a^2+b^2+c^2)|`

In the given equation,

`a = 3, b = -4, c = 12, d =-3`

`D =|(-3)/sqrt(3^2+(-4)^2+12^2)| = |(-3)/sqrt(9+16+144)| = |(-3)/sqrt169|`

`=>D = 3/13`


So, distance of the given plane from the origin is `3/13`.