3.12 In the first grade, more than 100 students took exams in Chinese, English, and mathematics. Among them, 92 passed the Chinese exam, 75 passed the English exam, and 65 passed the mathematics exam; 65 passed both Chinese and English exams, 54 passed both Chinese and mathematics exams, and 45 passed both English and mathematics exams. Find the number of students who passed all three subjects.

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The problem can be solved using the principle of inclusion-exclusion. Let's denote:
- \( A \) as the set of students who passed the Chinese exam,
- \( B \) as the set of students who passed the English exam,
- \( C \) as the set of students who passed the mathematics exam.

We are given:
- \( |A| = 92 \)
- \( |B| = 75 \)
- \( |C| = 65 \)
- \( |A \cap B| = 65 \)
- \( |A \cap C| = 54 \)
- \( |B \cap C| = 45 \)

We need to find \( |A \cap B \cap C| \), the number of students who passed all three subjects.

Using the principle of inclusion-exclusion for three sets, we have:
\[
|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|
\]

We know that the total number of students is more than 100, so:
\[
|A \cup B \cup C| > 100
\]

Substituting the given values:
\[
|A \cup B \cup C| = 92 + 75 + 65 - 65 - 54 - 45 + |A \cap B \cap C|
\]
\[
|A \cup B \cup C| = 168 - 164 + |A \cap B \cap C|
\]
\[
|A \cup B \cup C| = 4 + |A \cap B \cap C|
\]

Since \( |A \cup B \cup C| > 100 \):
\[
4 + |A \cap B \cap C| > 100
\]
\[
|A \cap B \cap C| > 96
\]

Since \( |A \cap B \cap C| \) must be an integer, the smallest possible value is 97. Therefore, the number of students who passed all three subjects is:
\[
\boxed{97}
\]
None

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None
