Answer: 10
P(AuBuC) = P(A) + P(B) + P(C) - P(AnB) - P(AnC) - P(BnC) + P(AnBnC)
Let $Y$ be the no. of persons who eat at least one item.
Y = 9 + 10 + 7 - [P(AnB) + P(AnC) + P(BnC)] + 5
[P(AnB) + P(AnC) + P(BnC)] = 31 - Y.
Now, these include the no. of persons who eat all 3 items thrice. So, excluding those we get, no. of persons who eat at least two items as
31 - Y - 2*5 = 21 - Y.
Minimum value of Y is 10 as 10 people eat fish. Is this possible? Yes.
Maximum value of Y is 21. Is this possible? No. Because 5 person eat all three item. So, the no. of persons eating at most 2 items = (9-5) + (10-5) + (7-5) = 11. And adding 5 we get 16 people who eat at least one item.
So, our required answer is $21-10 \geq X \geq 21 - 16 \implies 5 \leq X \leq 11.$