21 Can we take a ball, a tetrahedron, and a plane, such that every plane parallel to the chosen plane intersects the ball and the tetrahedron in sections of equal area?
21 can. Take a tetrahedron $ABCD$, with $E$ and $F$ being the midpoints of $AB$ and $CD$ respectively, and $EF$ being the common perpendicular segment of the skew lines $AB$ and $CD$. Given $AB=CD=a$, $EF=h$. Place the edge $CD$ on the plane $\alpha$, and make $AB \parallel \alpha$, at this time $EF \perp \alpha$. Take another sphere with diameter $h$, let it be tangent to the plane $\alpha$, and on the same side of the plane $\alpha$ as the tetrahedron $ABCD$.
Let $\beta$ be any plane parallel to $\alpha$, the distance between $\beta$ and $\alpha$ being $x (0<x<h)$, the sections of $\beta$ cutting the sphere and the tetrahedron are $S_{\odot}$ and $S_{\square}$ respectively. It is easy to know
$$
S_{\odot}=\pi x(h-x), S_{\square}=\left(\frac{a}{h}\right)^{2} x(h-x).
$$

Choose $\frac{a}{h}=\pi$.
