5. A plane divides the medians of the faces $A B C$, $A C D$ and $A B D$ of the tetrahedron $A B C D$ emanating from vertex $A$ in the ratios $1: 2, 1: 1$ and $1: 2$, counting from point $A$. Find the ratio of the volumes of the parts into which this plane divides the pyramid.
3.II.5. Let $ABCD$ be the given tetrahedron, and $AK, AM, AN$ be the medians of its faces $ABC, ACD, ABD$. Denote by $E, F$, and $G$ the points of intersection of the plane $PQR$ with the medians $AK, AM$, and $AN$, respectively (figure).

![](https://cdn.mathpix.com/cropped/2024_05_21_46613852391e49da1fa9g-071.jpg?height=459&width=579&top_left_y=1478&top_left_x=727)

Since $AE: EK = AG: GN$, it follows that $EG \parallel KN$. Similarly, $KN \parallel CD$. Therefore, $EG \parallel CD$, which means $EG \parallel (ACD)$. Consequently, the plane $PQR$, containing the line $EG$, intersects the plane $ACD$ along the line $QR$, which is parallel to $EG$, and thus $QR \parallel CD$. Since $QR$ passes through the midpoint $F$ of the median $AM$, the points $Q$ and $R$ are the midpoints of the segments $AC$ and $AD$.

Let the median $CT$ of triangle $ABC$ intersect the median $AK$ at point $S$, and the side $AB$ at point $T$ (figure).

![](https://cdn.mathpix.com/cropped/2024_05_21_46613852391e49da1fa9g-072.jpg?height=479&width=483&top_left_y=363&top_left_x=775)

Then $AS = \frac{2}{3} AK$ and, since $AE = \frac{1}{3} AK$, point $E$ is the midpoint of segment $AS$. Since $Q$ is the midpoint of $AC$, point $P$ is the midpoint of $AT$, so $AP: AB = 1: 4$. Therefore,

\[
\frac{V_{APQR}}{V_{ABCD}} = \frac{AP \cdot AQ \cdot AR}{AB \cdot AC \cdot AD} = \frac{1 \cdot 1 \cdot 1}{4 \cdot 2 \cdot 2} = \frac{1}{16}
\]

Thus,

\[
\frac{V_{APQR}}{V_{BCDPQR}} = \frac{1}{15}
\]