1. Let $f(n) = n^2+6n+11$ be a function defined on positive integers. Find the sum of the first three prime values $f(n)$ takes on.

[i]Proposed by winnertakeover[/i]
1. We start with the function \( f(n) = n^2 + 6n + 11 \) and need to find the sum of the first three prime values it takes on.

2. The smallest possible value of \( f(n) \) is when \( n = 1 \):
   \[
   f(1) = 1^2 + 6 \cdot 1 + 11 = 1 + 6 + 11 = 18
   \]
   Since 18 is not a prime number, we need to find other values of \( n \) for which \( f(n) \) is prime.

3. To determine when \( f(n) \) is prime, we consider the congruence properties of \( f(n) \). Note that:
   \[
   f(n) = n^2 + 6n + 11 \equiv n^2 + 6n + 11 \pmod{6}
   \]
   Simplifying, we get:
   \[
   f(n) \equiv n^2 + 0 + 5 \equiv n^2 + 5 \pmod{6}
   \]
   For \( f(n) \) to be prime, \( f(n) \equiv \pm 1 \pmod{6} \). Thus:
   \[
   n^2 + 5 \equiv \pm 1 \pmod{6}
   \]
   This implies:
   \[
   n^2 \equiv -4 \pmod{6} \quad \text{or} \quad n^2 \equiv 2 \pmod{6}
   \]
   Since 2 is not a quadratic residue modulo 6, we only consider:
   \[
   n^2 \equiv 0 \pmod{6}
   \]
   Therefore, \( n \) must be divisible by 6.

4. We test the first few multiples of 6:
   - For \( n = 6 \):
     \[
     f(6) = 6^2 + 6 \cdot 6 + 11 = 36 + 36 + 11 = 83
     \]
     83 is a prime number.
   - For \( n = 12 \):
     \[
     f(12) = 12^2 + 6 \cdot 12 + 11 = 144 + 72 + 11 = 227
     \]
     227 is a prime number.
   - For \( n = 18 \):
     \[
     f(18) = 18^2 + 6 \cdot 18 + 11 = 324 + 108 + 11 = 443
     \]
     443 is a prime number.

5. Summing the first three prime values of \( f(n) \):
   \[
   83 + 227 + 443 = 753
   \]

The final answer is \(\boxed{753}\).