Let $S$ be area of the given parallelogram $ABCD$ and the points $E,F$ belong to $BC$ and $AD$, respectively, such that $BC = 3BE, 3AD = 4AF$. Let $O$ be the intersection of $AE$ and $BF$. Each straightline of $AE$ and $BF$ meets that of $CD$ at points $M$ and $N$, respectively. Determine area of triangle $MON$.
1. **Identify the given ratios and points:**
   - \( BC = 3BE \) implies \( BE = \frac{BC}{3} \).
   - \( 3AD = 4AF \) implies \( AF = \frac{3AD}{4} \).

2. **Determine the coordinates of points \( E \) and \( F \):**
   - Since \( E \) divides \( BC \) in the ratio \( 1:2 \), \( E \) is one-third of the way from \( B \) to \( C \).
   - Since \( F \) divides \( AD \) in the ratio \( 3:1 \), \( F \) is three-fourths of the way from \( A \) to \( D \).

3. **Find the intersection point \( O \) of \( AE \) and \( BF \):**
   - Use the section formula to find the coordinates of \( O \). Since \( E \) and \( F \) divide the sides in known ratios, we can use the section formula to find \( O \).

4. **Determine the points \( M \) and \( N \):**
   - \( M \) is the intersection of \( AE \) with \( CD \).
   - \( N \) is the intersection of \( BF \) with \( CD \).

5. **Calculate the lengths \( CM \) and \( DN \):**
   - Since \( \Delta EAB \sim \Delta EMC \), the ratio of the sides gives \( CM = 2AB \).
   - Similarly, \( DN = \frac{AB}{3} \).

6. **Calculate the length \( MN \):**
   \[
   MN = CM + DN + AB = 2AB + \frac{AB}{3} + AB = \frac{6AB}{3} + \frac{AB}{3} + \frac{3AB}{3} = \frac{10AB}{3}
   \]

7. **Determine the area of \( \Delta MON \):**
   - The area of \( \Delta MON \) can be found by comparing it to the area of \( \Delta OAB \).
   - Use the height ratios of \( \Delta OMN \) and \( \Delta OAB \).

8. **Calculate the area of \( \Delta MON \):**
   - The area of \( \Delta MON \) is proportional to the area of the parallelogram \( ABCD \).
   - Given the height ratios and the base \( MN \), the area of \( \Delta MON \) is:
   \[
   \text{Area of } \Delta MON = \frac{50}{39}S
   \]

The final answer is \(\boxed{\frac{50}{39}S}\).