Find the probability that no two among $A$, $B$, and $C$ are together when $12$ people are arranged in a circle There are $12$ people including A,B and C. They are arranged in a circle. Find the probability that no two among A, B and C are together.
I have solved problem where cases involving two person sitting adjacent to each other. But here there are $3$ people where no two among them sits adjacent, hence I am getting confused.

I think it is easier to find the probability where the condition is not satisfied. We have two cases to consider:
Case 1: A,B,C are adjacent. This is simply grouping A,B,C and changing their places among themselves with $3!$ and merging them with the remaining $9$ people. In this case, there are
$$3!\cdot(10-1)! = 3!\cdot9!$$
such arrangements.
Case 2: A,B are adjacent, C is not adjacent to neither A nor B. We can find this by grouping $4$ people. While grouping them, put A and B on the two middle places with $2!$ and for the side places, we can choose $2$ people from $9$ (Not $10$ because we don't want C to be one of these side places as it is already counted in case 1) and change their places among themselves with $\binom{9}{2}2! = 72$. Then we can take this group as one person and merge them with the remaining $8$ people. In this case, there are
$$2!\cdot 72\cdot(9-1)! = 72\cdot 2! \cdot 8! = 16\cdot9!$$
such arrangements. Notice that this is the same for the cases where A,C are adjacent B is non-adjacent to neither A nor C; and B,C are adjacent A is non-adjacent to neither B nor C. So, in total, we have 
$$6\cdot9!+3\cdot16\cdot9! = 54\cdot9!$$ 
arrangements that do not satisfy the condition. So the answer should be
$$\frac{(12-1)!-54\cdot9!}{(12-1)!} = \frac{11\cdot10\cdot9!-54\cdot9!}{11!} = \frac{56\cdot9!}{11!} = \frac{28}{55}$$
