We know that,$A\cup B\cup C=A+B+C-A\cap B-A\cap C-B\cap C+A\cap B\cap C$.
For $|A\cup B\cup C|$ to be minimum,$|A|,|B|,|C|,$$|A\cap B\cap C|$ has to be minimum and $|A\cap B|,|B\cap C|, |A\cap C|$ has to be maximum.
So let us assume that $B\cap C$={1,2,3,4,5,6,7,8}.So the elements 1,2,3,4,5,6,7,8 has to be in both $B$,$C$ for sure.
Now B can be of the form $B$={1,2,3,4,5,6,7,8....} which can be visualized as B={1,2,3,4,5,6,7,8} $\cup$ $X$.
where X is the set of remaining elements of $B$ apart from {1,2,3,4,5,6,7,8}.Here X can be finite or infinite.
Now since $|B|$ has to be minimum, $|X|$ has to be zero i.e.,$X$ has to be a empty set.
$B=${1,2,3,4,5,6,7,8} $\cup$ $X$$=${1,2,3,4,5,6,7,8}.
i.e.,$B$={1,2,3,4,5,6,7,8}
Thinking in same manner,$C$={1,2,3,4,5,6,7,8}.
Now coming to $|A\cap C|$$=7$.
since $C$={1,2,3,4,5,6,7,8}, we can assume $A\cap C$={1,2,3,4,5,6,7}
Hence we can assume $A$={1,2,3,4,5,6,7....}={1,2,3,4,5,6,7} $\cup$ $X$
where $X$ is the set of remaining elements of $A$ apart from {1,2,3,4,5,6,7,8}.
Now since $|A|$ has to be minimum, $|X|$ has to be zero i.e.,$X$ has to be a empty set.
$\therefore$ $A$={1,2,3,4,5,6,7}
Hence $A\cap B\cap C=${1,2,3,4,5,6,7}. and $|A\cap B\cap C|=7$.
$|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|A\cap C|-|B\cap C| +|A\cap B\cap C|$.
=7+8+8-7-7-8+7
=8
Here, $A\cap C$ can be any of {1,2,3,4,5,6,7} or {1,2,3,4,5,6,8} or {1,2,3,4,5,7,8} or {1,2,3,4,6,7,8} or {1,2,3,5,6,7,8} or{1,2,4,5,6,7,8} or {1,3,4,5,6,7,8} or {2,3,4,5,6,7,8}.
Just remove one element from $C$.That will give you $A\cap C$
