Binary search complexity In sorted array of numbers binary search gives us comlexity of O(logN).
How will the complexity change if we split array into 3 parts instead of 2 during search?

You can apply the master theorem directly using:
$$
T(n) = T\left( \frac{n}{3} \right) + O(1)
$$
But you can also think intuitively reasoning that if you search on a portion of the array that is 1/3 every time, and you finish when the array portion is of length 1 you need $k$ steps to complete the algorithm, with $k$ computed as:
$$
\frac{n}{3^k} = 1
$$
Therefore $k = log_3 n$, and $O(\log_3 n)$ is the complexity of the algorithm. 
The ternary search suggested by Clement C. solves a slightly different problem, where you actually search for a maximum or minimum and not for a specific, given, item. In that case your recursion goes on 2/3 of the array instead of 1/3. I remains true, in any case, that the result is different only by a multiplicative constant, because you can always change the base.
