A square-based (with base vertices $A, B, C, D$) equal-sided pyramid has its apex at $E$. Let $P$ be the point that divides the edge $AE$ in a 3:1 ratio, such that $EP:PA=3$, and let $Q$ be the midpoint of the edge $CE$! In what ratio does the plane passing through points $D, P$, and $Q$ divide the edge $BE$?
Let's denote the intersection point under investigation as $R$ and the center of the base as $T$. Since both $P$ and $Q$ lie in the diagonal plane $A C E$ of the pyramid, the line $P Q$ intersects the pyramid's $E T$ axis. Let's denote this intersection point as $S$. The segments $Q T$ and $P E$ are parallel; furthermore, $Q T$ is half, and $P E$ is three-quarters of the segment $A E$. Therefore, the triangles $S Q T$ and $S P E$ are similar, and in them, $S T: S E = Q T: P E = 2: 3$.

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Draw a line parallel to the $E T$ axis through the vertex $B$. This line lies in the diagonal plane $B D E$, so it intersects the line $D S$. Let's denote this intersection point as $V$. Due to the similarity of the triangles $D T S$ and $D B V$, we have $S T: B V = 1: 2$. The similarity of the triangles $R E S$ and $R B V$ gives us

$$
B R: R E = B V: S E = 4: 3
$$