If  the number of three element subsets drawn from the set $S=\left\{1,2,3,\dots .,29,30\right\}$ are such that the sum of the three elements is a multiple of 3 is P, then the value of  $\frac{P}{340}$ is …........

Each of the 30 integers can fall into following set having 10 elements in each 
To;{3,6,9,12,15,18,21,24,27,30} $\to 3K$
To;{1,4,7,10,13,16,19,22,25,28} $\to 3K+1$
To;{2,5,8,11,14,17,20,23,26,29} $\to 3K+2$
Consider a general subset {x,y,z} if S, so that x+y+z=3
Beacause multiple of 3 can be formed out of 3
Remainders in extectly four different ways 0+0=0,1+1+1,2+2+2 and 0+1+2.
$\Rightarrow$  Either x, y and z selected from same set  $(T_0,T_1,T_2)$ or exactly one of them lie in each of these sets  $(T_0,T_1,T_2)$.
 $\Rightarrow P=3{\times }^{10}{C}_3+{(}^{10}{C_1}^{10}{C_1}^{10}{C}_1)=1360$
 $\therefore \frac{P}{340}=4$