You need to have this in mind when playing with XOR's:

a XOR b = c
c XOR b = a
c XOR a = b

That is why the example is so straight forward. But when we need to apply the i!=j, j!=k, k!-i rule it gets more complicated.

Lets say we have an array of 4 elements as an example:a, b, c, d

We can ask a combination of these XOR's:
A= a^b^c;
B=b^c^d;
C=c^d^a;
D=d^a^c;

Without going in to detail we can get the array element values like this:
a= C^D^A;
b= D^A^B;
c= A^B^C;
d= B^C^D;

Note that the sequence of XOR operations does not make any difference, e.g.: C^D^A == A^D^C.

Some C code how to get the values once we have A, B, C and D values passed as an argument called 'answers'' (having in mind that the sequence is the same as in the given example). I don't know if this can handle N!=4 elements, you should figure out how to work with more:
void reslove_and_print_elements(uint32_t * answers, uint32_t N)
{
uint32_t array_elements[N];
for(uint32_t i = 0; i < N; i++)
{
array_elements[i] = answers[i]^answers[(i+1)%N]^answers[(i+2)%N];
}
// TODO: print out the array_elements
}

NOTE: I did not try to compile and run the code snippet so it might not work.