12.2.9 * Given a regular tetrahedron $ABCD$ with edge length 2, the sum of the areas of all sections obtained by planes equidistant from its four vertices is ( ).
(A) 4
(B) 3
(C) $\sqrt{3}$
(D) $3+\sqrt{3}$
There are two types of cross-sections:
(1) One side of the cross-section has one vertex, and the other side has three vertices, such as $\triangle B_{1} C_{1} D_{1}$ in Figure a, where $B_{1}$, $C_{1}$, and $D_{1}$ are the midpoints of edges $A B$, $A C$, and $A D$, respectively. There are four such cross-sections, each with an area of $\frac{\sqrt{3}}{4}$;
Figure a
Figure b
(2) Both sides of the cross-section have two vertices each, such as the quadrilateral $M N P Q$ in Figure b, where $M$, $N$, $P$, and $Q$ are the midpoints of edges $A B$, $B C$, $C D$, and $D A$, respectively. There are three such cross-sections, all of which are squares, each with an area of 1. In summary, the total area of all cross-sections is $3+\sqrt{3}$. Therefore, the answer is $D$.