`a, b, c` are positive integers froming an increasing `G.P`. whose common ratio is rational number, `b - a` is cube of natural number and `log_(6)a + log_(6)b + log_(6)c = 6` then `a + b + c` is divisible by

Options

A. `3`
B. `5`
C. `7`
D. `9`

Explanation

To solve the problem, we need to find the sum \( a + b + c \) where \( a, b, c \) are positive integers forming an increasing geometric progression (G.P.) with a rational common ratio, and given conditions about \( b - a \) and logarithmic relationships.

### Step-by-Step Solution:

1. **Understanding the G.P.**:
Since \( a, b, c \) are in G.P., we can express them in terms of \( a \) and the common ratio \( r \):
\[ b = ar \quad \text{and} \quad c = ar^2 \]
where \( r \) is a rational number.

2. **Using the logarithmic condition**:
We are given that:
\[ \log_6 a + \log_6 b + \log_6 c = 6 \]
Using the property of logarithms, we can combine these:
\[ \log_6 (abc) = 6 \]
This implies:
\[ abc = 6^6 \]

3. **Expressing \( abc \)**:
Substituting \( b \) and \( c \):
\[ abc = a \cdot ar \cdot ar^2 = a^3 r^3 \]
Therefore, we have:
\[ a^3 r^3 = 6^6 \]
Simplifying gives:
\[ (ar)^3 = 6^6 \]
Hence:
\[ ar = 6^2 = 36 \quad \text{(since } ar = b\text{)} \]

4. **Finding \( b \)**:
We already found \( b = 36 \).

5. **Finding \( a \) and \( c \)**:
We know \( b - a \) is a cube of a natural number. Let \( b - a = k^3 \) for some natural number \( k \):
\[ 36 - a = k^3 \implies a = 36 - k^3 \]
Since \( a \) must be positive, we have:
\[ 36 - k^3 > 0 \implies k^3 < 36 \]
The cubes less than 36 are \( 1^3 = 1, 2^3 = 8, 3^3 = 27 \). Thus, possible values for \( k \) are \( 1, 2, 3 \).

6. **Calculating possible values for \( a \)**:
- If \( k = 1 \): \( a = 36 - 1 = 35 \)
- If \( k = 2 \): \( a = 36 - 8 = 28 \)
- If \( k = 3 \): \( a = 36 - 27 = 9 \)

7. **Finding \( c \)**:
Using \( c = ar^2 \) and \( r = \frac{b}{a} \):
- For \( a = 35 \): \( r = \frac{36}{35} \) and \( c = 36 \cdot \left(\frac{36}{35}\right) = \frac{1296}{35} \) (not an integer)
- For \( a = 28 \): \( r = \frac{36}{28} = \frac{9}{7} \) and \( c = 36 \cdot \left(\frac{9}{7}\right)^2 = \frac{36 \cdot 81}{49} = \frac{2916}{49} \) (not an integer)
- For \( a = 9 \): \( r = \frac{36}{9} = 4 \) and \( c = 36 \cdot 4 = 144 \) (integer)

8. **Final values**:
Thus, the valid values are:
\[ a = 9, \quad b = 36, \quad c = 144 \]

9. **Calculating \( a + b + c \)**:
\[ a + b + c = 9 + 36 + 144 = 189 \]

10. **Finding divisibility**:
We check for divisibility of \( 189 \) by \( 3, 5, 7, 9 \):
- \( 189 \div 3 = 63 \) (divisible)
- \( 189 \div 5 \) (not divisible)
- \( 189 \div 7 = 27 \) (divisible)
- \( 189 \div 9 = 21 \) (divisible)

### Conclusion:
The sum \( a + b + c = 189 \) is divisible by \( 3, 7, \) and \( 9 \).