`a`, `b`, `c` are positive integers formaing an incresing `G.P.` and `b-a` is a perfect cube and `log_(6)a+log_(6)b+log_(6)c=6`, then `a+b+c=`

Options

A. `100`
B. `111`
C. `122`
D. `189`

Explanation

To solve the problem, we need to find the values of \( a \), \( b \), and \( c \) given that they are positive integers forming an increasing geometric progression (G.P.), \( b - a \) is a perfect cube, and \( \log_6 a + \log_6 b + \log_6 c = 6 \).

### Step-by-Step Solution:

1. **Understanding the Geometric Progression**:
Since \( a \), \( b \), and \( c \) are in G.P., we can express them in terms of a common ratio \( r \):
\[ b = ar \quad \text{and} \quad c = ar^2 \]
Here, \( a \) is the first term, \( b \) is the second term, and \( c \) is the third term.

2. **Using the Logarithmic Equation**:
The equation given is:
\[ \log_6 a + \log_6 b + \log_6 c = 6 \]
Using the properties of logarithms, this can be rewritten as:
\[ \log_6 (abc) = 6 \]
Therefore, we have:
\[ abc = 6^6 \]

3. **Substituting the Values of \( a \), \( b \), and \( c \)**:
Substituting the expressions for \( b \) and \( c \):
\[ a \cdot (ar) \cdot (ar^2) = 6^6 \]
This simplifies to:
\[ a^3 r^3 = 6^6 \]
Taking the cube root of both sides:
\[ ar = 6^2 = 36 \]
Thus, we have:
\[ b = 36 \]

4. **Finding \( a \) and \( c \)**:
Since \( b = ar \), we can express \( a \) as:
\[ a = \frac{36}{r} \]
And for \( c \):
\[ c = ar^2 = 36r \]

5. **Finding \( b - a \)**:
We know that:
\[ b - a = 36 - \frac{36}{r} = 36 \left(1 - \frac{1}{r}\right) \]
This must be a perfect cube. Let’s denote \( k = 36 \left(1 - \frac{1}{r}\right) \), and we need \( k \) to be a perfect cube.

6. **Finding Possible Values of \( r \)**:
The possible integer values for \( r \) that keep \( a \), \( b \), and \( c \) as positive integers are the factors of 36 (excluding 1 to maintain the increasing order):
- Possible values for \( r \): 2, 3, 4, 6, 9, 12, 18, 36.

7. **Testing Values of \( r \)**:
We will check these values to see which one makes \( b - a \) a perfect cube:
- For \( r = 2 \):
\[ b - a = 36(1 - \frac{1}{2}) = 36 \cdot \frac{1}{2} = 18 \quad \text{(not a perfect cube)} \]
- For \( r = 3 \):
\[ b - a = 36(1 - \frac{1}{3}) = 36 \cdot \frac{2}{3} = 24 \quad \text{(not a perfect cube)} \]
- For \( r = 4 \):
\[ b - a = 36(1 - \frac{1}{4}) = 36 \cdot \frac{3}{4} = 27 \quad \text{(perfect cube, } 3^3\text{)} \]

8. **Calculating \( a \), \( b \), and \( c \)**:
- For \( r = 4 \):
\[ a = \frac{36}{4} = 9, \quad b = 36, \quad c = 36 \cdot 4 = 144 \]

9. **Finding \( a + b + c \)**:
Now, we can find:
\[ a + b + c = 9 + 36 + 144 = 189 \]

### Final Answer:
\[ \boxed{189} \]