Out of a group of $21$ persons, $9$ eat vegetables, $10$ eat fish and $7$ eat eggs. $5$ persons eat all three. How many persons eat at least two out of the three dishes?
My approach:- $N(A∪B∪C)=N(A)+N(B)+N(C)−N(A∩B)−N(A∩C)−N(B∩C)+N(A∩B∩C)$
$21=9+10+7−N(A∩B)−N(A∩C)−N(B∩C)+5$
$N(A∩B)+N(A∩C)+N(B∩C)=10$
Now the LHS has counted $N(A∩B∩C)$ three times, so I will remove it two times as:-
Number of persons eating at least two dishes $=N(A∩B+B∩C+A∩C ) - 2*N(A∩B∩C) =10-2*5 =0$
Now it contradicts the questions that there are $5$ eating all three dishes.
Is this anything wrong in my approach?
As far as I can see there is something wrong with your approach, but also the question does not have enough information to give a definite answer.
Your mistake is to assume that $N(A\cup B\cup C)=21$. However presumably the $21$ people may include some who eat neither vegetables nor fish nor eggs: if there are $n$ such people then you should have $$N(A\cup B\cup C)=21-n\ .$$ If you now follow your method you should get the number of people you want also to be $n$. And since $n$ is not known, the problem cannot be answered. In fact as you point out, the answer must be at least $5$, so we know that $n\ge5$. You can find solutions by trial and error with $n=5,6,7,8,9,10$, so any of these is a possible answer.