30 people were invited to a dinner party. If 17 of the guest

You can simplify the problem by considering below scenario.
"30 people, invited to a dinner party. Inviter has three different types of meal pases- A, B, C. He is going to distribute these in such that no one gets multiple passes of same type and least number of get all three passes."

In this question, answer is whatever number that is surplus of 60 i.e. (30 X 2).

Q 1) People=30
A= 17
B= 15
C= 20
Answer is 0. but to chose the least option in the given choices, I would choose 2 i.e. option A.

Q 2) People =30
A= 25
B=21
C=20
least number of people, who get 3 passes=
= (25+21+20)- 60
= 6
Answer is 6. Option D.


Below is one more similar analogy that will help to understand the problems.
There are 30 boxes placed in a row. You have three types of marbles- A, B & C. You are told place such that, no box hold 2 marbles of same type and least number of boxes should hold all types of marbles.

Consider below image of puts and marbles places inside.