If `a, b, c` are three natural numbers in AP such that `a + b + c=21` and if possible number of ordered triplet `(a, b, c)` is `lambda`, then the value of `(lambda -5)` is

Explanation

To solve the problem, we need to find the number of ordered triplets \((a, b, c)\) of natural numbers that are in arithmetic progression (AP) and satisfy the equation \(a + b + c = 21\).

### Step-by-Step Solution:

1. **Understanding Arithmetic Progression**:
Since \(a, b, c\) are in AP, we can express \(b\) in terms of \(a\) and \(c\):
\[ b = \frac{a + c}{2} \]
This implies:
\[ 2b = a + c \]

2. **Substituting into the Sum Equation**:
We know from the problem that:
\[ a + b + c = 21 \]
Substituting \(b\) from the AP condition into this equation:
\[ a + \frac{a + c}{2} + c = 21 \]
Multiplying through by 2 to eliminate the fraction:
\[ 2a + (a + c) + 2c = 42 \]
Simplifying this gives:
\[ 3a + 3c = 42 \]
Dividing by 3:
\[ a + c = 14 \]

3. **Finding \(b\)**:
From the previous step, we can find \(b\):
\[ b = \frac{a + c}{2} = \frac{14}{2} = 7 \]

4. **Expressing \(c\) in terms of \(a\)**:
Since \(a + c = 14\), we can express \(c\) as:
\[ c = 14 - a \]

5. **Finding the Range for \(a\)**:
Since \(a, b, c\) are natural numbers, \(a\) must be a natural number such that both \(a\) and \(c\) remain positive:
\[ a \geq 1 \quad \text{and} \quad c = 14 - a \geq 1 \]
This gives:
\[ 14 - a \geq 1 \implies a \leq 13 \]
Therefore, \(a\) can take values from 1 to 13.

6. **Counting the Ordered Triplets**:
For each value of \(a\) from 1 to 13, there is a corresponding value of \(c\) given by \(c = 14 - a\), and \(b\) is always 7. Thus, the ordered triplets \((a, b, c)\) can be written as:
\[ (1, 7, 13), (2, 7, 12), (3, 7, 11), \ldots, (13, 7, 1) \]
This gives us 13 valid ordered triplets.

7. **Final Calculation**:
Let \(\lambda\) be the number of ordered triplets, which we found to be 13. We need to find:
\[ \lambda - 5 = 13 - 5 = 8 \]

### Conclusion:
The value of \((\lambda - 5)\) is \(8\).