For $$$a$$$ $$$XOR$$$ $$$b$$$ $$$XOR$$$ $$$c = 0$$$, it must be the case that each bit is either set in $$$0$$$ or $$$2$$$ among the three integers. Let us consider the set $$$S$$$ of the bit positions (starting with $$$0$$$ for least significant bit) which are set in exactly $$$2$$$ of these. We can now write: $$$N = 2*(\sum_{x \in S} 2^{x})$$$.
Clearly, if $$$N$$$ is odd no solution may exist. If $$$N$$$ is even, the above equation gives a representation of $$$\frac{N}{2}$$$ as a sum of distinct powers of $$$2$$$. This representation must be equivalent to the binary representation of $$$\frac{N}{2}$$$.
Therefore, we have uniquely recovered the set of bits which must be set in exactly two of the outputs. Now, we only need to distribute them among the three numbers. First of all, if $$$|S| = 1$$$, there can be no answer (all three outputs are required to be positive). Otherwise, let $$$i$$$ be the smallest member of $$$S$$$, then the following triple $$$(a, b, c) = (2^{i}, \frac{N}{2} - 2^{i}, \frac{N}{2})$$$ is the required answer. (This is obtained by greedily minimising $$$a$$$, followed by greedily minimising $$$b$$$.)
Time Complexity: $$$O(logN)$$$ per test case.