Problem 1398

1. **Identify the properties of the rhombus $ABCD$:**
- Since $ABCD$ is a rhombus, all sides are equal: $AB = BC = CD = DA$.
- The diagonals $AC$ and $BD$ intersect at right angles and bisect each other.
- Given $\angle DAB = 60^\circ$, we can infer that $\angle ABC = 120^\circ$ because the sum of the angles in a rhombus is $360^\circ$.

2. **Locate the intersection point $P$:**
- The diagonals $AC$ and $BD$ intersect at point $P$, which is the center of the rhombus.

3. **Assume $R$ is on $AB$:**
- Without loss of generality, let $R$ be on $AB$. Since $PQRS$ is a rhombus, the diagonals of $PQRS$ must be perpendicular and bisect each other at $P$.

4. **Determine the positions of $Q$ and $S$:**
- Since $PQRS$ is a rhombus and $R$ is on $AB$, one of $Q$ or $S$ must be on $AD$ and the other on $BC$. This ensures that the diagonals of $PQRS$ are perpendicular and bisect each other at $P$.

5. Analyze the parallelism and midpoint properties:
- The line segment $PR$ must be parallel to $AD$ and $BC$ because the diagonals of $PQRS$ are perpendicular and bisect each other at $P$.
- Therefore, $R$ must be the midpoint of $AB$.

6. **Verify the equilateral triangle $BPR$:**
- Since $R$ is the midpoint of $AB$, $BPR$ forms an equilateral triangle with $\angle BPR = 60^\circ$.
- Given that the diagonals of $PQRS$ are perpendicular, $B$ must be one of the vertices of $PQRS$, and the midpoint of $AD$ must be the other vertex.

7. Conclusion:
- Since $R$ is on $AB$, and $B$ is one of the vertices of $PQRS$, exactly one of $Q, R, S$ lies on one of $A, B, C, D$.

$\blacksquare$