If `f(2-x)=f(2+x)` and `f(7-x)=f(7+x)` and `f(0)=0`. If the minimum number of roots of `f(x)=0` where `0lexle100` is `lamda` then `lamda//3` equals

Explanation

To solve the problem, we need to analyze the given functional equations and the conditions provided.

### Step-by-step Solution:

1. **Understanding the Functional Equations**:
We have two functional equations:
- \( f(2 - x) = f(2 + x) \) (Equation 1)
- \( f(7 - x) = f(7 + x) \) (Equation 2)

2. **Analyzing Equation 1**:
From Equation 1, we can deduce that the function is symmetric about \( x = 2 \). This means that if \( x \) is replaced by \( 2 + k \) or \( 2 - k \), the function will yield the same value.

3. **Analyzing Equation 2**:
Similarly, from Equation 2, we can deduce that the function is symmetric about \( x = 7 \).

4. **Finding the Periodicity**:
If we replace \( x \) in Equation 1 with \( x + 5 \), we get:
\[ f(2 - (x + 5)) = f(2 + (x + 5)) \]
Simplifying gives:
\[ f(-3 - x) = f(7 + x) \]
This shows that \( f(-3 - x) \) is equal to \( f(7 + x) \).

5. **Establishing Periodicity**:
From the above, we can conclude that:
\[ f(x) = f(x + 10) \]
This indicates that the function \( f(x) \) is periodic with a period of 10.

6. **Evaluating \( f(0) = 0 \)**:
Since \( f(0) = 0 \), and due to the periodicity, we can find other roots:
- \( f(10) = f(0) = 0 \)
- \( f(20) = f(0) = 0 \)
- Continuing this, we find:
- \( f(30) = 0 \)
- \( f(40) = 0 \)
- \( f(50) = 0 \)
- \( f(60) = 0 \)
- \( f(70) = 0 \)
- \( f(80) = 0 \)
- \( f(90) = 0 \)
- \( f(100) = 0 \)

7. **Counting the Roots**:
The roots of \( f(x) = 0 \) in the interval \( [0, 100] \) are:
- \( 0, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100 \)
This gives us a total of 11 roots.

8. **Finding \( \lambda \)**:
Thus, the minimum number of roots \( \lambda = 11 \).

9. **Calculating \( \lambda // 3 \)**:
Finally, we need to compute \( \lambda // 3 \):
\[ \lambda // 3 = 11 // 3 = 3 \]

### Final Answer:
\[ \lambda // 3 = 3 \]