If `a,b,c gt1` then `Delta=|(log_(a)(abc),log_(a)b,log_(a)c),(log_b(abc),1,log_(b)c),(log_(c)(abc),log_(c)b,1)|` equals

Options

A. 0
B. `log_(a)b+log_(b)c+log_(c)a`
C. `log_(abc)(a+b+c)`
D. None of these

Answer

To solve the problem, we need to evaluate the determinant given by:

\[ \Delta = \begin{vmatrix} \log_a(abc) & \log_a b & \log_a c \\ \log_b(abc) & 1 & \log_b c \\ \log_c(abc) & \log_c b & 1 \end{vmatrix} \]

### Step 1: Rewrite the logarithms using natural logarithms
We can express the logarithms in terms of natural logarithms (ln) using the change of base formula:

\[ \log_a x = \frac{\ln x}{\ln a} \]

Thus, we can rewrite each term in the determinant:

- \(\log_a(abc) = \frac{\ln(abc)}{\ln a} = \frac{\ln a + \ln b + \ln c}{\ln a}\)
- \(\log_a b = \frac{\ln b}{\ln a}\)
- \(\log_a c = \frac{\ln c}{\ln a}\)
- \(\log_b(abc) = \frac{\ln(abc)}{\ln b} = \frac{\ln a + \ln b + \ln c}{\ln b}\)
- \(\log_b c = \frac{\ln c}{\ln b}\)
- \(\log_c(abc) = \frac{\ln(abc)}{\ln c} = \frac{\ln a + \ln b + \ln c}{\ln c}\)
- \(\log_c b = \frac{\ln b}{\ln c}\)

Substituting these into the determinant gives us:

\[ \Delta = \begin{vmatrix} \frac{\ln(abc)}{\ln a} & \frac{\ln b}{\ln a} & \frac{\ln c}{\ln a} \\ \frac{\ln(abc)}{\ln b} & 1 & \frac{\ln c}{\ln b} \\ \frac{\ln(abc)}{\ln c} & \frac{\ln b}{\ln c} & 1 \end{vmatrix} \]

### Step 2: Factor out the common terms from each row
We can factor out \(\frac{1}{\ln a}\), \(\frac{1}{\ln b}\), and \(\frac{1}{\ln c}\) from the respective rows:

\[ \Delta = \frac{1}{\ln a \cdot \ln b \cdot \ln c} \begin{vmatrix} \ln(abc) & \ln b & \ln c \\ \ln(abc) & \ln b & \ln c \\ \ln(abc) & \ln b & \ln c \end{vmatrix} \]

### Step 3: Analyze the determinant
Notice that the first and second rows (and the first and third rows) of the determinant are identical. According to the properties of determinants, if two rows (or two columns) are identical, the value of the determinant is zero.

### Conclusion
Thus, we conclude that:

\[ \Delta = 0 \]

### Final Answer
The value of the determinant \(\Delta\) is \(0\).

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