Out of $3n$ consecutive integers, three are selected at random. The chance that their sum is divisible by $3$ is 

Let $3n$ consecutive integers be 

$N+1,N+2,N+3,\dots \dots \dots \dots ..,N+3n$(starting with the integer $N)$

We write these $3n$ numbers in 3 rows as following;

$\begin{array}{l}N+1,N+4,N+7,\dots \dots \dots \dots \dots ,N+3n-2\\ N+2,N+5,N+8,\dots \dots \dots \dots \dots ,N+3n-1\\ N+3,N+6,N+9,\dots \dots \dots \dots .,N+3n\end{array}$

The sum of three selected number will be divisible by $3$ it either all three belong to the same row or all three belong to different rows. So, the favourable no. of cases 

$=3\left({ }^n{C}_3\right)+\left({ }^n{C}_1\right)\left({ }^n{C}_1\right)\left({ }^n{C}_1\right)$

$=\frac{3n(n-1)(n-2)}{3!}+n^3=\frac{3n^3-3n^2+2n}{2}$

Also, the total no of cases

${=}^{3n}{C}_3=\frac{3n(3n-1)(3n-2)}{3!}=\frac{n(3n-1)(3n-2)}{2!}$

$\therefore$ Required probability 

   $=\frac{\frac{3n^3-3n^2+2n}{2}}{\frac{n(3n-1)(3n-2)}{2}}=\frac{3n^2-3n+2}{(3n-1)(3n-2)}$