What does it mean to find the distance from the origin to a plane in $\mathbb R^3$? In do Carmo, one exercise gives a plane in $\mathbb R^3$, $ax +by +cz+d = 0$, and tells us to show that $|d|/\sqrt{a^2 + b^2 + c^2}$ measures the distance from the plane to the origin.
However, this seems a bit ambiguous since we don't know what the plane actually is.
By distance, does he mean minimal distance?

For the plane $ax+by+cz + d = 0$,
the normal vector is: $$\hat n = <a,b,c>$$
The vector from the plane to any arbitrary point is: $$\hat v = <(x-x_0),(y-y_0),(z-z_0)>$$
If we consider the origin in particular, $(x_0,y_0,z_0) = (0,0,0)$ so, $$\hat v = <(x),(y),(z)>$$
The MINIMUM distance from the origin to the plane is the projection of v onto n: $$\frac{|\hat n\cdot \hat v|}{|\hat n|}$$ Plugging in the vectors and computing the dot product yields:$$\frac{ax+by+cz}{\sqrt {a^2+b^2+c^2}}$$
From the plane's equation, $-d = ax+by+cz$ 
Thus the minimum distance from the origin to a plane is given by: $$\frac{|d|}{\sqrt {a^2+b^2+c^2}}$$
