**Given Information:**

- There are three positive integers a, b, and c.
- The highest common factor (HCF) of each possible pair of a, b, and c is 11.
- The product of all three numbers is 13310.

**Approach:**

We need to find all possible sets of {a, b, c} that satisfy the given conditions.

To solve this problem, we can use the concept of prime factorization and the properties of HCF.

**Step 1: Prime Factorization of 13310**

Let's find the prime factorization of 13310:
13310 = 2 × 5 × 7 × 7 × 11

**Step 2: HCF of Each Possible Pair**

We know that the highest common factor (HCF) of each possible pair of a, b, and c is 11. Therefore, each number in the set {a, b, c} must have a factor of 11.

**Step 3: Possible Sets of {a, b, c}**

Since each number in the set {a, b, c} must have a factor of 11, we can consider the prime factorization of 13310.

- The factor 11 can be included in any one of the three numbers.
- The remaining prime factors (2, 5, 7, 7) can be distributed among the three numbers in various ways, keeping in mind that each number must have a factor of 11.

Let's consider the possible distributions of the remaining prime factors:

- Case 1: One number has factor 11, and the other two numbers have the remaining prime factors (2, 5, 7, 7).
- Case 2: Two numbers have factor 11, and the remaining number has the remaining prime factors (2, 5, 7, 7).
- Case 3: All three numbers have factor 11, and each number has one or more of the remaining prime factors (2, 5, 7, 7).

**Step 4: Counting Possible Sets**

To count the possible sets, we need to consider the number of ways we can distribute the remaining prime factors.

- Case 1: One number has factor 11, and the other two numbers have the remaining prime factors (2, 5, 7, 7).
- In this case, we have 3 choices for the number with factor 11 and 2 choices for each remaining prime factor.
- So, the total number of sets in this case is 3 × 2 × 2 = 12.

- Case 2: Two numbers have factor 11, and the remaining number has the remaining prime factors (2, 5, 7, 7).
- In this case, we have 3 choices for the two numbers with factor 11 and 3 choices for each remaining prime factor.
- So, the total number of sets in this case is 3 × 3 × 3 = 27.

- Case 3: All three numbers have factor 11, and each number has one or more of the remaining prime factors (2, 5, 7, 7).
- In this case, each number can have any combination of the remaining prime factors.
- Since there are 4 remaining prime factors,