6. The section of the rectangular parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ by a plane passing through its vertex $D$ intersects the lateral edges $A A_{1}, B B_{1}$, and $C C_{1}$ at points $K, M$, and $N$, respectively. Find the ratio of the volume of the pyramid with vertex at point $P$ and base $D K M N$ to the volume of the given parallelepiped, if it is known that point $P$ divides the edge $D D_{1}$ in the ratio $D P: P D_{1}=m: n$.
4.I.6. Let $a, b$, and $c$ denote the lengths of the edges $A D, D C$, and $D D_{1}$ of the given parallelepiped, and let $V$ be its volume. According to the property of parallel planes, $K M \| D N$ and $M N \| K D$, so $D K M N$ is a parallelogram (figure).

![](https://cdn.mathpix.com/cropped/2024_05_21_46613852391e49da1fa9g-077.jpg?height=525&width=548&top_left_y=1091&top_left_x=754)

Therefore, the plane $K P N$ divides the pyramid $P D K M N$ into two equal-volume pyramids. Thus, the volume of the pyramid $P D K M N$ is twice the volume of the pyramid $N D K P$. Drop perpendiculars $N E$ and $K F$ to the line $D D_{1}$. Since the planes $A D D_{1}$ and $C D D_{1}$ are perpendicular, $N E \perp (A D D_{1})$. Therefore,

$$
V_{P D K M N}=2 V_{N D K P}=2 \cdot \frac{1}{3} S_{D K P} \cdot N E=2 \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot D P \cdot K F \cdot N E
$$

Since $D P=\frac{m}{m+n} c, K F=a$, and $N E=b$, we finally obtain that

$$
V_{P D K M N}=\frac{m}{3(m+n)} a b c=\frac{m}{3(m+n)} V
$$