10. (20 points) As shown in Figure 2, in the tetrahedron $ABCD$,

Figure 2

it is known that the edges $AB, AC, AD$ are mutually perpendicular, $BC = CD = 10$, and $BD = 12$. A plane $\alpha_{A}$ parallel to the base $BCD$ is constructed such that the distance between $\alpha_{A}$ and the base $BCD$ is 1. Similarly, construct the other three planes $\alpha_{B}$, $\alpha_{C}$, and $\alpha_{D}$. Find the volume of the small tetrahedron formed by the intersection of the four planes.
10. As shown in Figure 6, let the sections $\alpha_{B}, \alpha_{C}, \alpha_{D}$ intersect at a point $A'$. Similarly, define points $B', C', D'$.

Thus, the tetrahedron $A' B' C' D'$ is similar to the tetrahedron $A B C D$.

Let the similarity ratio be $k$, and the volumes of tetrahedra $A B C D$ and $A' B' C' D'$ be $V$ and $V'$, respectively.
Let $A B = a, A C = b, A D = c$. Then $a^{2} + b^{2} = b^{2} + c^{2} = 100, c^{2} + a^{2} = 144$.
Solving these, we get $a = c = 6 \sqrt{2}, b = 2 \sqrt{7}$.
Therefore, $V = \frac{1}{6} a b c = 24 \sqrt{7}$.
Also, $S_{\triangle B C D} = 6 \sqrt{10^{2} - 6^{2}} = 48$, so the distance from point $A$ to the plane $B C D$ is
$$
A F = \frac{3 V}{S_{\triangle B C D}} = \frac{3 \sqrt{7}}{2}.
$$

Then, $\cos \angle C A F = \frac{A F}{A C} = \frac{3}{4}$.
Therefore, the distance from point $A'$ to the plane $B C D$ is
$$
\left[(6 - \sqrt{2}) \frac{2 \sqrt{7}}{6} - 1\right] \times \frac{3}{4} = \frac{3 \sqrt{7}}{2} - \frac{\sqrt{14}}{4} - \frac{3}{4}.
$$

Thus, the distance from point $A'$ to the plane $B' C' D'$ is $\frac{3 \sqrt{7}}{2} - \frac{\sqrt{14}}{4} - \frac{3}{4} - 1 = \frac{3 \sqrt{7}}{2} - \frac{\sqrt{14}}{4} - \frac{7}{4}$. Therefore, $k = \frac{\frac{3 \sqrt{7}}{2} - \frac{\sqrt{14}}{4} - \frac{7}{4}}{\frac{3 \sqrt{7}}{2}} = \frac{1}{6}(6 - \sqrt{7} - \sqrt{2})$. Hence, $V' = k^{3} V = \left(\frac{6 - \sqrt{7} - \sqrt{2}}{6}\right)^{3} \times 24 \sqrt{7} = \frac{\sqrt{7}}{9}(6 - \sqrt{7} - \sqrt{2})^{3}$.