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 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. 
