Let $a_k,b_k,c_k$ be integers, for $k=1,2,3...,n$ and let $f(x)$ be the number of ordered triples $(A,B,C)$ of subsets (not necessarily non-empty) of $S$ with $A \cup B \cup C=S=\{1,2,3...,n\}$ such that $$\sum_{i \in S\backslash A}a_i+\sum_{i \in S\backslash B}b_i+\sum_{i \in S\backslash C}c_i \equiv x \pmod{3}$$ Given that $f(0)=f(1)=f(2)$, prove there exists $i\in S$ such that $3 \vert a_i+b_i+c_i$.
My approach was to first consider $$\sum_{k=0}^{2}f(k)\varepsilon^k=\sum_{\\ A \cup B \cup C=S}\varepsilon^{\sum_{i \in S\backslash A}a_i+\sum_{i \in S\backslash B}b_i+\sum_{i \in S\backslash C}c_i}=0$$ where $\varepsilon=e^{\frac{2\pi i}{3}}$.
I tried to come up with a product formula for the right-hand side and got
$$\sum_{\\ A \cup B \cup C=S}\varepsilon^{\sum_{i \in S\backslash A}a_i+\sum_{i \in S\backslash B}b_i+\sum_{i \in S\backslash C}c_i}= \prod_{i=1}^{n}(\varepsilon^{a_i}+\varepsilon^{b_i}+\varepsilon^{c_i}+\varepsilon^{a_i+b_i}+\varepsilon^{a_i+c_i}+\varepsilon^{b_i+c_i}+1)$$
Here each term in the product counts the exponent in the different cases as $i$ is in at least one of $A,B,C$.
$1$. The $\varepsilon^{a_i+b_i}$ term is in the case where $i \in C$ but $i \notin A,B$
$2$. The $\varepsilon^{a_i}$ term is in the case where $i\in B,C$ but $i\notin A$.
$3$. The $1$ term is when $i\in A,B,C$.
Rest of the terms being symmetric.
I'm unsure where to proceed from here as there is no $\varepsilon^{a_i+b_i+c_i}$ term in the product and I'm not even certain if the formula is correct. Any help would be much appreciated.
[With deference to the authors, I'm not fully confident about this, so please point out any concerns that you have.]
Since there are $ 7^n$ ordered triples $(A, B, C)$, which means that it is impossible to get $ f(0) = f(1) = f(2)$, and thus we can draw any conclusions that we want
In particular, we can show that for any set of integers
$$\varepsilon^{a_i}+\varepsilon^{b_i}+\varepsilon^{c_i}+\varepsilon^{a_i+b_i}+\varepsilon^{a_i+c_i}+\varepsilon^{b_i+c_i} + 1 \neq 0,$$ and hence their product cannot be 0.
Perhaps, what the authors intended was also that $ A \cap B \cap C = \emptyset$, in which case there are $6^n$ ordered triples and it is not (yet) impossible to get $f(0) = f(1) = f(2)$. In this scenario, we have
$$ 0 = \prod_{i=1}^n (\varepsilon^{a_i}+\varepsilon^{b_i}+\varepsilon^{c_i}+\varepsilon^{a_i+b_i}+\varepsilon^{a_i+c_i}+\varepsilon^{b_i+c_i}), $$
so one of the terms must be equal to 0.
Can you complete the proof from there?