find 3 non repeating numbers in an array using xor and constant extra space

let XOR = xor of all the numbers in the array let say

then the ith bit of xor will be 1 (set)

only if either 

1. All the 3 non repeating numbers have ith bit set.

2. 1 non repeating number has ith bit set.

and the ith bit of xor will be 0 (unset)

only if either

1. All the 3 non repeating numbers have ith bit unset.

2. 1 non repeating number has ith bit unset.

we make two arrays named set and unset

CASE 1. ( ith bit is set )

if the ith bit of xor is set then we check if the set array has ith element not equal to xor value,

then its a non repeating element. Thats because, 

in that case only 2. point satisfies as stated above which is only the number.

CASE 2. ( ith bit is unset )

in this case too only 2. point of the unset statement says that it will be only one element else it will be xor of all the elements.