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Ivan Izmestiev
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As j.c. mentioned in his answer, the average distance between parallel supporting planes is better known as the mean width. More generally, one can take a convex body in $\mathbb{R}^n$ and consider the average $k$-volume of its projections to $k$-dimensional linear subspaces (the case $k=1$ is the mean width). These quantities are called quermassintegrals, and they are proportional to the coefficients in the Steiner formula for the volume of the parallel body. Your question is about the case $n=3$, $k=1$ of this theorem.

To prove it, start with $n=3$ and $k=2$. This is Cauchy's theorem which says that the area of the boundary of the convex body is one quarter of its average projection area: $$ Area(\partial K) = \frac1{\pi} \int\limits_{\xi} Area(pr_\xi K)\, d\xi, $$ where the integral is taken over all unit vectors $\xi$, and $pr_\xi K$ is the projection of $K$ to $\xi^\perp$. The proof is very nice, just integrate the projection area of each face individually.

Now replace $K$ in the Cauchy formula by the $K_\epsilon$, the $\epsilon$-neighborhood of $K$ and apply the Steiner formula to both sides. The LHS is easy to compute by decomposing the boundary of $K_\epsilon$ into flats, pieces of cylinders, and pieces of spheres: $$ Area(\partial K_\epsilon) = Area(\partial K) + \epsilon\sum_e \ell_e(\pi-\delta_e) + \epsilon^2 \cdot 4\pi. $$ For the RHS observe that $pr_\xi(K_\epsilon) = (pr_\xi K)_\epsilon$ and expand the integral in a similar way: $$ Area(pr_\xi K_\epsilon) = Area(pr_\xi K) + \epsilon L(\partial pr_\xi K) + \epsilon^2 \cdot \pi. $$ Now integrating and comparing the coefficients at $\epsilon$ we obtain $$ \sum_e \ell_e(\pi-\delta_e) = \frac1\pi \int_\xi L(\partial pr_\xi K)\, d\xi =\frac1\pi \int_\xi \frac12 \int_{\eta \in \xi^\perp} L(pr_{\xi\oplus\eta}K)\, d\eta d\xi, $$ where we applied the Crofton formula relating the length of a curve to the average length of its projections. It remains to transform the right hand side: $$ \frac1\pi \int_\xi \frac12 \int_{\eta \in \xi^\perp} L(pr_{\xi\oplus\eta}K)\, d\eta d\xi = \int_\zeta L(pr_{\zeta^\perp}K)\, d\zeta, $$ and we get the integral of lengths of projections of $K$ to lines.

This proof comes from 1920's, I guess, and the procedure is called Kubota's recursion. For exact references see Notes for Section 4.5 of

Schneider, Rolf, Convex bodies: the Brunn-Minkowski theory, Encyclopedia of Mathematics and Its Applications. 44. Cambridge: Cambridge University Press. xiii, 490 p. (1993). ZBL0798.52001.

Ivan Izmestiev
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