First and foremost, it can be proven that $$$(a \oplus b) + (b \oplus c) + (a \oplus c)$$$ is always even, for all possible non-negative values of $$$a$$$, $$$b$$$ and $$$c$$$.
ProofFirstly, $$$a \oplus b$$$ and $$$a+b$$$ have the same parity, since $$$a + b = a \oplus b + 2 \cdot (a \text{&} b) $$$. Therefore, $$$(a \oplus b) + (b \oplus c) + (a \oplus c)$$$ has the same parity as $$$(a+b)+(b+c)+(a+c)=2 \cdot (a+b+c)$$$.
Therefore, if $$$n$$$ is even, one possible solution is $$$a=0$$$, $$$b=0$$$ and $$$c=\frac{n}{2}$$$. In this case, $$$(a \oplus b) + (b \oplus c) + (a \oplus c)= 0+\frac{n}{2}+\frac{n}{2}=n$$$. Otherwise, there are no solutions.
Time complexity per testcase: $$$O(1)$$$.