Example 6 For any set $S$, let $|S|$ denote the number of elements in set $S$, and let $n(S)$ denote the number of subsets of set $S$, including the empty set and set $S$ itself. If $A$, $B$, and $C$ are three sets that satisfy the condition $n(A) + n(B) + n(C) = n(A \cup B \cup C)$ and $|A| = |B| = 100$, then the minimum value that $|A \cap B \cap C|$ can take is ( ).
(A) 96
(B) 97
(C) 98
(D) 99
(E) 100
(42nd American High School Mathematics Examination)
Solution: Since a set containing $k$ elements has $2^{k}$ subsets, from $n(A)+n(B)+n(C)=n(A \cup B \cup C)$, we get $2^{100}+2^{100}+2^{|C|}=2^{|A \cup B \cup C|}$.
$$
\therefore 1+2^{|C|-101}=2^{|A \cup B \cup C|-101}
$$

It is easy to see that $1+2^{|C|-101}$ is a number greater than 1 and a power of 2.
$$
\therefore|C|=101 \text {. }
$$

Thus, $|A \cup B \cup C|=102$.
By the principle of inclusion-exclusion, we have
$$
\begin{array}{l}
|A \cap B \cap C| \\
=|A|+|B|+|C|+|A \cup B \cup C| \\
-|A \cup B|-|B \cup C|-|C \cup A| \text {. } \\
\because|A \cup B|,|B \cup C|,|C \cup A| \leqslant 102 \text {, } \\
\therefore|A \cap B \cap C| \geqslant 100+100+101+102 \\
-102-102-102 \\
=97 \text {. } \\
\end{array}
$$

The following example shows that 97 can be achieved:
$$
\begin{array}{l}
A=\{1,2, \cdots, 97,98,99,100\}, \\
B=\{1,2, \cdots, 97,98,101,102\}, \\
C=\{1,2, \cdots, 97,99,100,101,102\},
\end{array}
$$
$\therefore|A \cap B \cap C|$'s minimum possible value is 97. Therefore, the answer should be (B).