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?
 A: By distance, the author presumably implies minimal distance to the plane, which is achieved by finding a vector that is perpendicular to the plane, and then using that vector to find the distance between the origin and the plane.
More precisely, given a general plane of equation ax+by+cz+d=0, we can find the perpendicular vector (a,b,c).
The reason as to why the minimal distance is the formula provided can be seen in a much better exposition here
A: Yes he does mean minimal distance. I'm working on a proof for the $\mathbb R ^n$ case, and will update once this is done. 
Consider the analog in $\mathbb R^2$, distance $= |c| / (\sqrt{a^2+b^2})$, for a line $ax+by+c=0$.
Example: $x+y+1=0 \Rightarrow y=-1-x$. Clearly the minimum (Euclidean) distance from the origin is $\sqrt{2}$.
A: 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: $$\frac{|d|}{\sqrt {a^2+b^2+c^2}}$$
A: A different way using optimization:
\begin{align}
\min\  & x^2+y^2+z^2  & \text{(squared distance to the origin)}\\
 & ax+by+cz+d=0 & \text{(the point belongs to the plane)}
\end{align}
The Lagrangian is 
$$
\mathcal{L}(x,y,z)= x^2+y^2+z^2 + \lambda(ax+by+cz+d)
$$
Solving $\nabla\mathcal{L}=0$ (where the gradient is taken w.r.t. to $x,y,z$) you get:
$$
\nabla\mathcal{L} =2\left(\begin{array}{c}x\\y\\z\end{array}\right)+\lambda\left(\begin{array}{c}a\\b\\c\end{array}\right)=0
$$
Now observe that like the point $(x,y,z)$ belongs to the plane you have:
$$\langle \left(\begin{array}{c}x\\y\\z\end{array}\right),\left(\begin{array}{c}a\\b\\c\end{array}\right) \rangle =-d
$$
Thus
$$
\langle \nabla\mathcal{L}, \left(\begin{array}{c}a\\b\\c\end{array}\right) \rangle  = -2d+\lambda(a^2+b^2+c^2)=0
$$
which gives you the multiplier value: $$\lambda = \frac{2d}{a^2+b^2+c^2}$$
Going back to the initial equation $\nabla\mathcal{L}=0$ you get the point $(x_\star,y_\star,z_\star)$ that realizes the minimum:
$$
\left(\begin{array}{c}x_\star\\y_\star\\z_\star\end{array}\right) =\frac{d}{a^2+b^2+c^2}\left(\begin{array}{c}a\\b\\c\end{array}\right)
$$
Taking the norm you get its distance to the origin:
$$
\sqrt{x_\star^2+y_\star^2+z_\star^2}=\frac{|d|}{\sqrt{a^2+b^2+c^2}}
$$
To answer to your question, yes this formula is the minimal distance to a point $(x,y,z)$ in your plane to the origin. 
