Determine all triplets $(a,b,c)$ of real numbers such that sets $\{a^2-4c, b^2-2a, c^2-2b \}$ and $\{a-c,b-4c,a+b\}$ are equal and $2a+2b+6=5c$. In every set all elements are pairwise distinct
1. We are given the equations:
   \[
   2a + 2b + 6 = 5c
   \]
   and
   \[
   \{a^2 - 4c, b^2 - 2a, c^2 - 2b\} = \{a - c, b - 4c, a + b\}
   \]
   where the elements in each set are distinct.

2. According to the set equality, we have:
   \[
   (a^2 - 4c) + (b^2 - 2a) + (c^2 - 2b) = (a - c) + (b - 4c) + (a + b)
   \]
   Simplifying the right-hand side:
   \[
   (a - c) + (b - 4c) + (a + b) = a - c + b - 4c + a + b = 2a + 2b - 5c
   \]
   Therefore, we have:
   \[
   a^2 + b^2 + c^2 - 2a - 2b - 4c = 2a + 2b - 5c
   \]

3. Combining this with the given equation \(2a + 2b + 6 = 5c\), we substitute \(5c = 2a + 2b + 6\) into the equation:
   \[
   a^2 + b^2 + c^2 - 2a - 2b - 4c = -6
   \]
   Rearranging terms, we get:
   \[
   a^2 + b^2 + c^2 - 2a - 2b - 4c + 6 = 0
   \]
   This can be rewritten as:
   \[
   (a - 1)^2 + (b - 1)^2 + (c - 2)^2 = 0
   \]

4. Since the sum of squares is zero, each square must individually be zero:
   \[
   (a - 1)^2 = 0 \implies a = 1
   \]
   \[
   (b - 1)^2 = 0 \implies b = 1
   \]
   \[
   (c - 2)^2 = 0 \implies c = 2
   \]

5. We now verify that the triplet \((a, b, c) = (1, 1, 2)\) satisfies the original conditions:
   - For the equation \(2a + 2b + 6 = 5c\):
     \[
     2(1) + 2(1) + 6 = 5(2) \implies 2 + 2 + 6 = 10 \implies 10 = 10
     \]
     This is true.
   - For the set equality:
     \[
     \{a^2 - 4c, b^2 - 2a, c^2 - 2b\} = \{1^2 - 4(2), 1^2 - 2(1), 2^2 - 2(1)\} = \{-7, -1, 2\}
     \]
     \[
     \{a - c, b - 4c, a + b\} = \{1 - 2, 1 - 4(2), 1 + 1\} = \{-1, -7, 2\}
     \]
     Both sets are \(\{-7, -1, 2\}\), which are equal.

The final answer is \( \boxed{ (1, 1, 2) } \).