This problem involves calculating when two people, starting from different points on an equilateral triangle and traveling at different speeds, will first meet. Let's solve this step-by-step.

Given:

• The triangle ABC is an equilateral triangle.
• The speeds of the two people are 30 km/hr and 20 km/hr, starting from points A and B respectively, moving in clockwise direction.

Objective:

Determine where the two people will first meet.

Solution:

1. Calculate the relative speed of the two people as they are moving in the same direction:
   • Person at A moves at 30 km/hr.
   • Person at B moves at 20 km/hr.
   • Relative speed = 30 km/hr - 20 km/hr = 10 km/hr.
2. Since they meet when the faster person (from A) catches up with the slower person (from B), we need to calculate when the faster person will complete one full lap and also position themselves at the starting point of the slower person.
3. As they meet when either completes one full lap of the triangle, let's assume the side length of the triangle is \( s \). Therefore, the perimeter is \( 3s \).
4. The time taken to complete one lap by the faster person moving at 30 km/hr is:
   • Time = \( \frac{3s}{30} \) hours.
5. The slower person will also take the same time but being slower, will revisit the starting point naturally or meet at similar multiple of timings.
6. Thus, they first meet again at their starting points simultaneously after the distance covered by each is exactly equal to the perimeter.

Therefore, the two people meet for the first time at point A.