108. The volume of the tetrahedron \(ABCD\) is 5. A plane is drawn through the midpoints of the edges \(AD\) and \(BC\), intersecting the edge \(CD\) at point \(M\). The ratio of the length of segment \(DM\) to the length of segment \(CM\) is \(2/3\). Calculate the area of the section of the tetrahedron by the specified plane, given that the distance from this plane to vertex \(A\) is 1.
108. Let $K$ and $L$ be the midpoints of edges $AD$ and $BC$, and let $N$ and $P$ be the points of intersection of the plane with lines $AB$ and $AC$ (Fig. 16). We need to find the ratios $\frac{|PA|}{|PC|}$ and $\frac{|PK|}{|PM|}$. Draw $KQ$ and $AR$ parallel to $DC$, with $Q$ being the midpoint of $AC$.

$$
|AR|=|DM|, \quad \frac{|PA|}{|PC|}=\frac{|AR|}{|MC|}=\frac{|DM|}{|MC|}=\frac{2}{3}
$$

$$
\frac{|PK|}{|PM|}=\frac{|KQ|}{|MC|}=\frac{|DC|}{2|MC|}=\frac{5}{6}
$$

Next, we find

$$
\begin{gathered}
\frac{|AN|}{|NB|}=\frac{2}{3}, \quad \frac{|PN|}{|PL|}=\frac{4}{5} \\
\frac{V_{PAKN}}{V_{ABCD}}=\frac{|PA| \cdot|AK| \cdot|AN|}{|AC| \cdot|AD| \cdot|AB|}=\frac{2}{5}
\end{gathered}
$$

i.e., $V_{PAKN}=2$. Since the height dropped from $A$ to $PNK$ is $1$, and $S_{PNK}=6$,

$$
\frac{S_{PML}}{S_{PNK}}=\frac{|PK| \cdot|PN|}{|PM| \cdot|PL|}=\frac{3}{2}, \quad S_{PML}=9
$$

Thus, the area of the section will be $S_{PML}-S_{PNK}=3$.
![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-070.jpg?height=342&width=924&top_left_y=674&top_left_x=165)

Fig. 16.