Sets $A, B$, and $C$ satisfy $|A| = 92$, $|B| = 35$, $|C| = 63$, $|A\cap B| = 16$, $|A\cap C| = 51$, $|B\cap C| = 19$. Compute the number of possible values of$ |A \cap  B \cap C|$.
1. Let \( |A \cap B \cap C| = x \). We need to find the possible values of \( x \).

2. Using the principle of inclusion-exclusion for three sets, we have:
   \[
   |A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|
   \]
   Substituting the given values:
   \[
   |A \cup B \cup C| = 92 + 35 + 63 - 16 - 51 - 19 + x
   \]
   Simplifying the expression:
   \[
   |A \cup B \cup C| = 104 + x
   \]

3. Since \( |A \cup B \cup C| \) must be a non-negative integer, we have:
   \[
   104 + x \geq 0 \implies x \geq -104
   \]
   However, this constraint is not useful since \( x \) must be a non-negative integer.

4. Next, consider the individual intersections:
   - For \( |A \cap B| = 16 \):
     \[
     |A \cap B| = |(A \cap B) \setminus C| + |A \cap B \cap C| = 16 - x
     \]
     Since \( |(A \cap B) \setminus C| \geq 0 \):
     \[
     16 - x \geq 0 \implies x \leq 16
     \]

   - For \( |A \cap C| = 51 \):
     \[
     |A \cap C| = |(A \cap C) \setminus B| + |A \cap B \cap C| = 51 - x
     \]
     Since \( |(A \cap C) \setminus B| \geq 0 \):
     \[
     51 - x \geq 0 \implies x \leq 51
     \]

   - For \( |B \cap C| = 19 \):
     \[
     |B \cap C| = |(B \cap C) \setminus A| + |A \cap B \cap C| = 19 - x
     \]
     Since \( |(B \cap C) \setminus A| \geq 0 \):
     \[
     19 - x \geq 0 \implies x \leq 19
     \]

5. Combining these constraints, we have:
   \[
   x \leq \min(16, 51, 19) = 16
   \]

6. Now, consider the constraint from the total number of elements in \( A \):
   \[
   |A| = |A \setminus (B \cup C)| + |A \cap B| + |A \cap C| - |A \cap B \cap C|
   \]
   Substituting the given values:
   \[
   92 = |A \setminus (B \cup C)| + 16 + 51 - x
   \]
   Simplifying:
   \[
   92 = |A \setminus (B \cup C)| + 67 - x
   \]
   \[
   |A \setminus (B \cup C)| = 25 + x
   \]
   Since \( |A \setminus (B \cup C)| \geq 0 \):
   \[
   25 + x \geq 0 \implies x \geq -25
   \]
   This constraint is not useful since \( x \) must be a non-negative integer.

7. Similarly, consider the constraint from the total number of elements in \( B \):
   \[
   |B| = |B \setminus (A \cup C)| + |A \cap B| + |B \cap C| - |A \cap B \cap C|
   \]
   Substituting the given values:
   \[
   35 = |B \setminus (A \cup C)| + 16 + 19 - x
   \]
   Simplifying:
   \[
   35 = |B \setminus (A \cup C)| + 35 - x
   \]
   \[
   |B \setminus (A \cup C)| = x
   \]
   Since \( |B \setminus (A \cup C)| \geq 0 \):
   \[
   x \geq 0
   \]

8. Finally, consider the constraint from the total number of elements in \( C \):
   \[
   |C| = |C \setminus (A \cup B)| + |A \cap C| + |B \cap C| - |A \cap B \cap C|
   \]
   Substituting the given values:
   \[
   63 = |C \setminus (A \cup B)| + 51 + 19 - x
   \]
   Simplifying:
   \[
   63 = |C \setminus (A \cup B)| + 70 - x
   \]
   \[
   |C \setminus (A \cup B)| = x - 7
   \]
   Since \( |C \setminus (A \cup B)| \geq 0 \):
   \[
   x - 7 \geq 0 \implies x \geq 7
   \]

9. Combining all the constraints, we have:
   \[
   7 \leq x \leq 16
   \]

10. Therefore, the number of possible values for \( x \) is:
    \[
    16 - 7 + 1 = 10
    \]

The final answer is \(\boxed{10}\)