Say I have one sphere and one plane
$$x^2+y^2+z^2 = 1$$ $$z = 2$$
We can easily calculate closest distance between these. It is easy exercise. Maybe first exercise in differential geometry or something like that. But is there some sense in which we can calculate "total closeness" of the objects? Like integrating over the objects calculating all distances and boil it down to one real number somehow.
I suggest to consider the orthogonal distance from a point of the sphere to the plan instead of the (non-orthogonal) distance to all points of the plan. So, the mean distance is not infinite.
If we consider the surface of the sphere, the orthogonal distance from the point $x,y,z$ to the plan $x=2$ is : $\big|2\pm\sqrt{1-y^2-z^2}\big|$. The sign depends on which hemisphere the point is located.
The mean distance is : $$\frac{\int_{-1}^1\int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}}\left(2+\sqrt{1-y^2-z^2}\right)dy\:dz+\int_{-1}^1\int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}}\left(2-\sqrt{1-y^2-z^2}\right)dy\:dz}{\int_{-1}^1\int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}} dy\:dz+\int_{-1}^1\int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}} dy\:dz}=\frac{8\pi}{4\pi}=2$$ Thus with this definition of the mean distance, the result is $2$. This was obvious because de mean distance of two symmetric points $(x,y,z)$ and $(-x,y,z)$ is $2$ and this is true for all couples of points of the sphere.
With the same definition and for the same raison the mean distance of all points inside the volume of the sphere will be found $=2$ by a triple integral.
Until the plan doesn't intersect the sphere the mean orthogonal distance is equal to the distance from the center of the sphere to the plan. If the plan intersect the sphere the calculus is more difficult. It should be interesting to compute the double and the triple integral of the absolute value $\big|2\pm\sqrt{1-y^2-z^2}\big|$ which is different from $\big(2\pm\sqrt{1-y^2-z^2}\big)$.
I can’t post a comment yet, but a possibly relevant and helpful piece of information is a classical theorem of Jacobi in Riemannian Geometry roughly stating that a geodesic segment minimizes the arc length with respect to neighboring curves with the same ends if it has no conjugate points. Calculating a geodesic requires solving a specific differential equation of degree two. For more information, see for example https://www.ime.usp.br/~gorodski/teaching/mat5771/ch5.pdf.