21. The volume of tetrahedron $ZABC$ is 5. A plane is drawn through the midpoint $K$ of edge $ZA$ and the midpoint $P$ of edge $BC$, intersecting edge $ZC$ at point $M$, such that $ZM: MC = 2: 3$. If the distance from vertex $A$ to the plane is 1, find the area of the section.
21. Let the section intersect $AB$ at $L$. By the general property 20 of a tetrahedron, we have $\frac{A K}{K Z} \cdot \frac{Z M}{M C} \cdot \frac{C P}{P B} \cdot \frac{B L}{L A}=1$, which means $\frac{B L}{L A}=\frac{3}{2}$. Connect $M A, M L, A P$, and let $S_{\triangle A B C}=S$. Draw the height $h$ of the tetrahedron $Z A B C D$ from point $Z$, then $\frac{1}{3} S h=5$. Draw the height of the tetrahedron $M A L P$ from point $M$ as $\frac{3}{5} h$, then $V_{M A P}=\frac{1}{3} S_{\triangle A I P} \cdot \frac{3}{5} h=\frac{1}{3} \cdot\left(\frac{2}{5} \cdot \frac{1}{2} S_{\triangle A B C}\right) \cdot \frac{3}{5} h=\frac{3}{25} \cdot \frac{1}{3} S h=\frac{3}{5}$.
Let $S_{\triangle Z A B}=S_{1}$, and draw the height of the tetrahedron $C Z A B$ from point $C$ as $h_{1}$, then $V_{M A K L}=\frac{1}{3} \cdot\left(\frac{1}{2} \cdot \frac{2}{5} S_{1}\right) \cdot \frac{2}{5} h_{1}=\frac{2}{25} \cdot \frac{1}{3} S_{1} h_{1}=$
$$
\begin{array}{l}
\frac{2}{5}, V_{A K M L L}=V_{M L P}+V_{M M L L}=1 . \\
\text { Also } V_{A K M P L}=\frac{1}{3} S_{K M P L} \cdot 1, \text { so } S_{\text {KMPL }}=3 \text { is the answer. }
\end{array}
$$