Solution:

The answer is (C). Let us select the people one at a time. At the beginning we have no constraints, so we can choose the first person in 32 ways. We can now choose the second person in 28 ways, excluding from the 32 the person already selected, the two adjacent to them, and the one facing them. However, we cannot consider them all together, because depending on how we choose the second person, the third can be chosen in different numbers of ways.

- If we choose the second person among the two seats adjacent to the one diametrically opposite our first choice, this excludes only one seat, in addition to the three already blocked before and the two already taken. Indeed, of the two adjacent seats, one was already blocked (diametrically opposite the first) and only one is now blocked, while the one diametrically opposite it had already been excluded since it is adjacent to the first. So we can choose the third among $32-1-3-2=26$ seats, for a total of $32 \cdot 2 \cdot 26$ choices for this first case.

- If we choose the second person among the two seats near the two adjacent to our first choice, we exclude the seat just taken, one of the two adjacent seats (the other was already blocked), and the diametrically opposite seat. So we can choose the third among $32-1-3-3=25$ seats, for a total of $32 \cdot 2 \cdot 25$ combinations.

- If we choose the second person among the 24 remaining seats, once those of the two previous cases have been excluded, this choice excludes 4 seats (the one chosen, the two adjacent, and the opposite one), for a total of $32 \cdot 24 \cdot 24$ combinations.

In total we then have $32\left(24^{2}+2 \cdot 25+2 \cdot 26\right)$ possible choices, to be divided by $3 !=6$ since we have taken into account the order in the choice of the people.