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: $N(A\cup B\cup C)=N(A)+N(B)+N(C)-N(A\cap B)-N(A\cap C)-N(B\cap C)+N(A\cap B\cap C)$

$21=9+10+7-N(A\cap B)-N(A\cap C)-N(B\cap C)+5$

$N(A\cap B)+N(A\cap C)+N(B\cap C)=10$

Now the LHS has counted $N(A\cap B\cap C)$ three times, so I will remove it two times as:-

Number of persons eating at least two dishes $=N(A\cap B+B\cap C+A\cap C)-2\ast N(A\cap B\cap C)=10-2\ast 5=0$

Now it contradicts the questions that there are 5 eating all three dishes.

Is this anything wrong in my approach?

My approach: $N(A\cup B\cup C)=N(A)+N(B)+N(C)-N(A\cap B)-N(A\cap C)-N(B\cap C)+N(A\cap B\cap C)$

$21=9+10+7-N(A\cap B)-N(A\cap C)-N(B\cap C)+5$

$N(A\cap B)+N(A\cap C)+N(B\cap C)=10$

Now the LHS has counted $N(A\cap B\cap C)$ three times, so I will remove it two times as:-

Number of persons eating at least two dishes $=N(A\cap B+B\cap C+A\cap C)-2\ast N(A\cap B\cap C)=10-2\ast 5=0$

Now it contradicts the questions that there are 5 eating all three dishes.

Is this anything wrong in my approach?