I solved this a while back. It's a bit challenging, but mostly due to the fact that the clues aren't really completely determining the solution, so you can't deduce each square like you can in e.g. sudoku. You have to make some guesses and then backtrack if they're wrong.
FWIW, I found that you can either deterministically deduce a cell or prove that the choice is irrelevant for at least two of the conditions. In this case, you can fill it with the solution that gives you the most flexibility for the third.
