`log_aN. log_bN+log_cN. log_bN+log_aN. log_cN`is equals to

Options

A. `(log_(a)N. log_(b)N. log_(c )N)/(log_(abc)N)`
B. `(log_(abc)N)/(log_(a)N. log_(b)N. log_(c )N)`
C. `(log_(N)abc)/(log_(N)a. log_(N)b. log_(N)c)`
D. none of these

Explanation

To solve the expression \( \log_a N \cdot \log_b N + \log_c N \cdot \log_b N + \log_a N \cdot \log_c N \), we will use the logarithmic base switch rule and properties of logarithms.

### Step-by-Step Solution:

1. **Apply the Logarithmic Base Switch Rule**:
The logarithmic base switch rule states that:
\[ \log_b N = \frac{1}{\log_N b} \]
Thus, we can rewrite each logarithm in the expression:
\[ \log_a N = \frac{1}{\log_N a}, \quad \log_b N = \frac{1}{\log_N b}, \quad \log_c N = \frac{1}{\log_N c} \]

2. **Rewrite the Expression**:
Substituting these into the original expression gives:
\[ \frac{1}{\log_N a} \cdot \frac{1}{\log_N b} + \frac{1}{\log_N c} \cdot \frac{1}{\log_N b} + \frac{1}{\log_N a} \cdot \frac{1}{\log_N c} \]

3. **Combine the Terms**:
Each term can be combined over a common denominator:
\[ = \frac{1}{\log_N a \cdot \log_N b} + \frac{1}{\log_N c \cdot \log_N b} + \frac{1}{\log_N a \cdot \log_N c} \]
The common denominator for all three terms is \( \log_N a \cdot \log_N b \cdot \log_N c \).

4. **Finding the Numerator**:
The numerator becomes:
\[ \frac{\log_N c + \log_N a + \log_N b}{\log_N a \cdot \log_N b \cdot \log_N c} \]

5. **Use the Logarithmic Product Rule**:
By the logarithmic product rule, we know:
\[ \log_N (abc) = \log_N a + \log_N b + \log_N c \]
Thus, we can rewrite the numerator:
\[ = \frac{\log_N (abc)}{\log_N a \cdot \log_N b \cdot \log_N c} \]

6. **Final Expression**:
Therefore, the entire expression simplifies to:
\[ = \frac{\log_N (abc)}{\log_N a \cdot \log_N b \cdot \log_N c} \]

### Final Answer:
\[ \log_a N \cdot \log_b N + \log_c N \cdot \log_b N + \log_a N \cdot \log_c N = \frac{\log_N (abc)}{\log_N a \cdot \log_N b \cdot \log_N c} \]