HCF or GCD of 3,6 and 18 is 3. This is the first time they meet.

t=750/3=250

Answer C

A completes 1 lap in 750/3 = 250 s
B completes 1 lap in 750/6 = 125 s
C completes 1 lap in 750/18 = 125/3 s

Comparing time vs laps completed in increments of 125/3 sec,

t=1*125/3 sec, A = 1/6 lap, B = 1/3 lap, C = 1 lap
t=2*125/3 sec, A = 2/6 lap, B = 2/4 lap, C = 2 laps
...
t=6*125/3 sec, A = 1 lap, B = 2 laps, C = 6 laps

So, they meet every 6*125/3 = 250 sec.

Alternatively, using LCM logic --

A and B meet every 250 sec
A and C meet every 125 sec
B and C meet every 250 sec

So, LCM of 3 cases above = 250 sec.

Plug in the answer choice, start with the middle option i.e. C

In 250 sec A will complete 750m
In 250 sec B will complete 1500m
In 250 sec C will complete 4500m

Since, they are all multiples of length of track so all the runners will at the same point on the track. Try with 50,75 sec as well to confirm whether 250 sec is the least time in which they will be together.

OA:

When two people are running in the same direction the relative speed is a difference in speeds of the two people.

In this case A=3 B=6 C=18

So relative speed of B wrt A is 6-3 = 3m/s
Relative speed of A wrt to C is 18-3 =15m/s

Therefore relative distances will be:
B wrt A is 750/3 =250
C wrt to A 750/15 = 50

So they have to bridge this distance of 250 and 50 between them which is the LCM of 250 and 50 which is 250.

Another Method: Simply put, Runner A's time take to run one lap is 250
Runner B's time is 125s
and Runner C's time is 41.67s

We can notice that A=2B
and thet B=3C

So when 250 s elapse, they will be at their starting point. A will have completed one lap, B 2 laps, and C 6 laps