If the largest number is $N$, then there are
possibilities for the two smaller numbers. Summing the possibilities for the smallest two numbers when the largest is each of 11,12,13,14, and 15 gives a total of
possible solutions for all three numbers. An efficient means of choosing a solution from a number of possibilities is to discard approximately half the possibilities with each yes/no question. This allows the questioner to distinguish between $2^k$ possibilities in $k$ questions. (Note that if there are exactly $2^k$ possibilities to start with you need a question that discards exactly half the possibilities).
For your set of possibilities, the question
"Is the largest number either 14 or 15?" (This reduces the set of 335 possibilities into a set of 166, if the answer is "No", or 169 if the answer is "Yes").
divides the possible answers into two almost equal sets. Whatever the answer, you can use the same strategy to get an answer within no more than