This is a problem involving the principle of inclusion-exclusion in set theory.

We are given:

• Total students ($N$) = 35
• Students in Cricket ($|C|$) = 20
• Students in Basketball ($|B|$) = 15
• Students in Football ($|F|$) = 13
• Students in at least two competitions ($|C \cap B| + |B \cap F| + |F \cap C|$) = 9
• We need to find $|C \cap B \cap F|$, the number of students who participate in all three competitions.

Step-by-step solution:

1. Use the inclusion-exclusion principle: According to the inclusion-exclusion principle for three sets: $|C \cup B \cup F| = |C| + |B| + |F| - (|C \cap B| + |B \cap F| + |F \cap C|) + |C \cap B \cap F|$

2. Substitute known values:

   • Total students: $|C \cup B \cup F| = 35$
   • Students in at least two competitions: $|C \cap B| + |B \cap F| + |F \cap C| = 9$

   Substituting into the inclusion-exclusion formula: $35 = 20 + 15 + 13 - 9 + |C \cap B \cap F|$

3. Simplify: $35 = 48 - 9 + |C \cap B \cap F|$

   $35 = 39 + |C \cap B \cap F|$

   $|C \cap B \cap F| = 35 - 39 = -4$

However, this result is