Let $k$ be a positive integer. $12k$ persons have participated in a party and everyone shake hands with $3k+6$ other persons. We know that the number of persons who shake hands with every two persons is a fixed number. Find $k.$
1. **Counting pairs \((A, \{B, C\})\) in two ways:**

   - First, consider each pair \((A, B)\). For each pair, there are \(d\) different people who shake hands with both \(A\) and \(B\). The number of such pairs \((A, B)\) is given by \(\binom{12k}{2} = \frac{12k(12k-1)}{2}\). Therefore, the total number of pairs \((A, \{B, C\})\) is:
     \[
     d \times \binom{12k}{2} = d \times \frac{12k(12k-1)}{2}
     \]

   - Second, consider each person \(P\). Each person shakes hands with \(3k+6\) others. The number of pairs \((A, \{B, C\})\) where \(A\) shakes hands with both \(B\) and \(C\) is given by \(\binom{3k+6}{2} = \frac{(3k+6)(3k+5)}{2}\). Since there are \(12k\) people, the total number of such pairs is:
     \[
     12k \times \binom{3k+6}{2} = 12k \times \frac{(3k+6)(3k+5)}{2}
     \]

2. **Equating the two expressions:**

   \[
   d \times \frac{12k(12k-1)}{2} = 12k \times \frac{(3k+6)(3k+5)}{2}
   \]

   Simplifying, we get:
   \[
   d \times 12k(12k-1) = 12k (3k+6)(3k+5)
   \]

   Dividing both sides by \(12k\):
   \[
   d \times (12k-1) = (3k+6)(3k+5)
   \]

   Expanding the right-hand side:
   \[
   d \times (12k-1) = 9k^2 + 33k + 30
   \]

   Solving for \(d\):
   \[
   d = \frac{9k^2 + 33k + 30}{12k-1}
   \]

3. **Finding \(k\) such that \(d\) is an integer:**

   Let \(p(k) = 9k^2 + 33k + 30\) and \(q(k) = 12k - 1\). We need \(q(k)\) to divide \(p(k)\). 

   We check the divisibility condition:
   \[
   q(k) \mid (k+4)q(k) - 4p(k)
   \]

   Calculating \((k+4)q(k) - 4p(k)\):
   \[
   (k+4)(12k-1) - 4(9k^2 + 33k + 30)
   \]

   Expanding and simplifying:
   \[
   12k^2 + 48k - k - 4(9k^2 + 33k + 30) = 12k^2 + 48k - k - 36k^2 - 132k - 120
   \]
   \[
   = -24k^2 - 85k - 120
   \]

   We need \(12k-1\) to divide \(-24k^2 - 85k - 120\). For \(k > 3\), \(|12k-1| > |3k-44|\). Therefore, the only possible value for \(k\) is \(k = 3\).

4. **Verification for \(k = 3\):**

   Substituting \(k = 3\):
   \[
   d = \frac{9(3)^2 + 33(3) + 30}{12(3) - 1} = \frac{81 + 99 + 30}{36 - 1} = \frac{210}{35} = 6
   \]

   Since \(d\) is an integer, \(k = 3\) is a valid solution.

The final answer is \( \boxed{ k = 3 } \).