Suppose `a,b,c epsilon R` and `a,b,c gt 0`.
Let `Delta=|(loga, logb, logc),(log(7a),log(49b),log(343c)),(log(3a),log(9b),log(27c))|` then `Delta` is equals to

Options

A. 0
B. `-1`
C. 1
D. 30

Answer

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

\[ \Delta = \begin{vmatrix} \log a & \log b & \log c \\ \log(7a) & \log(49b) & \log(343c) \\ \log(3a) & \log(9b) & \log(27c) \end{vmatrix} \]

### Step 1: Expand the logarithms in the determinant

Using the property of logarithms, we can expand the logarithmic terms:

\[ \log(7a) = \log 7 + \log a, \quad \log(49b) = \log 49 + \log b = 2\log 7 + \log b, \quad \log(343c) = \log 343 + \log c = 3\log 7 + \log c \]

\[ \log(3a) = \log 3 + \log a, \quad \log(9b) = \log 9 + \log b = 2\log 3 + \log b, \quad \log(27c) = \log 27 + \log c = 3\log 3 + \log c \]

Thus, we can rewrite the determinant as:

\[ \Delta = \begin{vmatrix} \log a & \log b & \log c \\ \log 7 + \log a & 2\log 7 + \log b & 3\log 7 + \log c \\ \log 3 + \log a & 2\log 3 + \log b & 3\log 3 + \log c \end{vmatrix} \]

### Step 2: Simplify the determinant

Now, we can simplify the determinant by subtracting the first row from the second and third rows:

\[ \Delta = \begin{vmatrix} \log a & \log b & \log c \\ \log 7 & \log 7 & \log 7 \\ \log 3 & \log 3 & \log 3 \end{vmatrix} \]

### Step 3: Identify identical rows

Notice that the second and third rows are now identical:

\[ \Delta = \begin{vmatrix} \log a & \log b & \log c \\ \log 7 & \log 7 & \log 7 \\ \log 3 & \log 3 & \log 3 \end{vmatrix} \]

Since the second and third rows are identical, the value of the determinant is zero:

\[ \Delta = 0 \]

### Conclusion

Thus, the value of \(\Delta\) is:

\[ \Delta = 0 \]