We have given three real numbers $a,b,c$ s.t.
$(1) \;\; 2a + 2b + 6 = 5c$
and
$(2) \;\; \{a^2 - 4c, b^2 - 2a, c^2 - 2b\} = \{a - c, b - 4c, a + b\}$,
where the numbers in the each of the two sets in (2) are distinct.
According to (2)
$(a^2 - 4c) + (b^2 - 2a) + (c^2 - 2b) = (a - c) + (b - 4c) + (a + b)$,
i.e.
$(3) \;\; a^2 + b^2 + c^2 - 2a - 2b - 4c = 2a + 2b - 5c$.
Combining (1) and (3) we obtain
$a^2 + b^2 + c^2 - 2a - 2b - 4c = - 6$,
which is equivalent to
$(4) \;\; (a - 1)^2 + (b - 1)^2 + (c - 2)^2 = 0$.
Consequently $(a,b,c) = (1,1,2)$ by (4), a triple which satisfies satisfies equation (1) and give us the two equal sets $\{-7,-1,2\}$ and $\{-1,-7,2\}$ in (2).
Conclusion: The only triple of real numbers satisfying the conditions (1) and (2) is $(a,b,c) = (1,1,2)$.