$h+t+u=21$

From the set of digits ${0, 1, 2, ..., 9}$ , inspect that the largest 3 digits from the set will be subtracted and will get 12, 13, and 14 as the remaining sum of the two digits. So we get $(9, t, u)$ , $(8, t, u)$ , and $(7, t, u)$ .

• If $t+u=12$ , then the possible $(t, u)$ pairs are $(8, 4)$ , and $(9, 3)$ . However, $(9, 9, 3)$ cannot be included here since we are looking for distinct digits, so we use $(9, 8, 4)$ .

• If $t+u=13$ , then the possible $(t, u)$ are $(7, 6)$ , and $(8, 5)$ . Again, $(8, 8, 5)$ is not included so we use $(8, 7, 6)$ .

• If $t+u=14$ , then the possible $(t, u)$ is $(9, 5)$ only and we consider to use $(7, 9, 5)$ .

So by permutation, $(9, 8, 4)$ can form 3-digit numbers in $3!$ ways (that is in $6$ ways), $(8, 7, 6)$ can form 3-digit numbers in $3!$ ways also, the same goes to $(7, 9, 5)$ . Hence, we have 18 distinct 3-digit numbers that add up to 21.