SOLUTION: There are 120 members in a certain activity club 52 take up dancing 58 joining the debating and 64 learn swimming 14 learn dancing and debating 27 debating and swimming and 32 swim

We are given 3 sets: D (dancing) of 52 elements;  B (debating) of 58 elements and W (swimming) of 64 elements.


We are also given the number of elements in each in-pair intersections.


We also are told that the union of all three sets contains 120 elements.


In such situation the following formula works


    n(D U B U W) = n(D) + n(B) + n(W) - n(D n B) - n(D n W) - n(B n W) + n(D n B n W).     (*)


Substituting the given data, you get


    120 = 52 + 58 + 64 - 14 - 27 - 32 + n(D n B n W).


In this equation, the only term  n(D n B n W)  is unknown - exactly that under the question.


So, it is easy to get it from the last equation:


    n(D n B n W) = 120 - 52 - 58 - 64 + 14 + 27 + 32 = 19.


Answer.  19 members enjoy all three activities.

    It is totally clear to you why I add the first three addends in the formula (*).


    But when I added them, I counted twice every term in each in-pair intersection.


    Therefore, I subtracted the number of terms in each in-pair intersection.


    Next, when I added three first addends, I counted thrice each term in the triple intersection;

    and when I subtracted in-pair intersections, I canceled these terms thrice.

    Therefore, I must add the number of terms in the triple intersection one more time to restore the balance.