I will assume (to make the solution as clear as possible) that all the pool of candidates is 100 persons.


To make the intersection of different categories minimal, we need (obviously) to make their union maximal.


So, i will assume that 


    a)  72 D-candidates and 75 I-candidates together cover the the entire set of 100 candidates  ====>  

        then their intersection DI consists of  72 + 75 - 100 = 47 persons.



    b)  72 D-candidates and 60 H-candidates together cover the entire set of 100 candidates  ====>  

        then their intersection DH consists of  72 + 60 - 100 = 32 persons.



    c)  75 I-candidates and 60 H-candidates together cover the entire set of 100 candidates  ====>  

        then their intersection IH consists of  75 + 60 - 100 = 35 persons.



Thus I have 3 sub-sets D, I and H  of 72, 75 and 60 elements, respectively, such that

            3 their intersections are  DI of 47 elements;  DH of 32 elements;  and  IH of 35 elements.



I also know that the union  D U I U H  consists of 100 elements.


Now I will use well known formula


    |D U I U H| = D + I + H - DI - DH - IH + IDH      (*)


  (in this formula each symbol means the number of elements in the corresponding subset) = 

                = 72+ 75+ 60 - 47 - 32 - 37 + DIH.


It gives me an equation


   100 = 72 + 75 + 60 - 47 - 32 - 37 + DIH,


which gives me the solution to the problem


   IDH = 100 - (72 + 75 + 60 - 47 - 32 - 37) = 9.


Answer.  The  minimal number of elements in the intersection IDH is 9,

         which gives the answer to the problem of 9%.