Let's solve this problem:

L and M are running on a circular track with speeds of 5 km/h and 3 km/h, respectively, in the same direction. The question asks how many times they will meet before returning to the starting point together.

Step-by-Step Explanation:

1. Relative Speed: Since they are running in the same direction, the relative speed at which they will approach each other is the difference in their speeds: $\text{Relative Speed} = 5 \text{ km/h} - 3 \text{ km/h} = 2 \text{ km/h}$

2. Time to meet: Suppose the length of the track is $d$. They will meet when the faster person (L) covers the extra distance relative to the slower person (M) to complete one lap.

   The time taken for the first meeting is given by: $\text{Time to meet} = \frac{d}{\text{Relative Speed}} = \frac{d}{2} \text{ hours}$

3. Time to return to the starting point: L and M will return to the starting point when the time taken by both runners is the least common multiple (LCM) of their time to complete one lap.

   • Time taken by L to complete one lap: $\frac{d}{5}$ hours
   • Time taken by M to complete one lap: $\frac{d}{3}$ hours

   The LCM of $\frac{d}{5}$ and $\frac{d}{3}$ is: $\text{LCM} \left(\frac{d}{5}, \frac{d}{3} \right) = \frac{d}{\text{GCD}(5, 3)} \times \text{LCM}(5, 3) = d \times \frac{15}{1} = 15$

   Therefore, they will meet back at the starting point after 15 units of time.

4. Number of meetings: In $\frac{15d}{5} = 3d$ hours, they will meet a certain number of times. The total number of times they will meet can be calculated as: $\frac{\text{Time taken to return}}{\text{Time taken to meet}} = \frac{15}{2} = 7.5$ This means they will meet 7 times before returning to the starting point.

Thus, the correct answer is 7 meetings. However, since the given options are limited to 1, 2, 3, or "Cannot be determined", it seems the correct interpretation of the question might involve a conceptual constraint.