If `a gt 0, b gt 0, c gt0` are respectively the pth, qth, rth terms of a G.P., then the value of the determinant
`|(log a,p,1),(log b,q,1),(log c,r,1)|`, is

Options

A. 0
B. log (abc)
C. `-(p+q+r)`
D. none of these

Explanation

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

\[ D = \begin{vmatrix} \log a & p & 1 \\ \log b & q & 1 \\ \log c & r & 1 \end{vmatrix} \]

where \( a, b, c \) are the \( p \)-th, \( q \)-th, and \( r \)-th terms of a geometric progression (G.P.).

### Step 1: Express \( a, b, c \) in terms of the first term and common ratio
Assume \( a \) is the first term and \( r \) is the common ratio of the G.P. Then we can express \( a, b, c \) as follows:
- \( a = A r^{p-1} \)
- \( b = A r^{q-1} \)
- \( c = A r^{r-1} \)

where \( A \) is the first term of the G.P.

### Step 2: Substitute \( a, b, c \) into the determinant
Now, we substitute these expressions into the determinant:

\[ D = \begin{vmatrix} \log (A r^{p-1}) & p & 1 \\ \log (A r^{q-1}) & q & 1 \\ \log (A r^{r-1}) & r & 1 \end{vmatrix} \]

Using the logarithm property \( \log (xy) = \log x + \log y \), we can rewrite the determinant:

\[ D = \begin{vmatrix} \log A + (p-1) \log r & p & 1 \\ \log A + (q-1) \log r & q & 1 \\ \log A + (r-1) \log r & r & 1 \end{vmatrix} \]

### Step 3: Simplify the determinant
We can simplify the determinant by performing row operations. Subtract the third row from the first and second rows:

\[ D = \begin{vmatrix} \log A + (p-1) \log r - (\log A + (r-1) \log r) & p & 1 \\ \log A + (q-1) \log r - (\log A + (r-1) \log r) & q & 1 \\ \log A + (r-1) \log r & r & 1 \end{vmatrix} \]

This simplifies to:

\[ D = \begin{vmatrix} ((p - r) \log r) & p & 1 \\ ((q - r) \log r) & q & 1 \\ ((r-1) \log r) & r & 1 \end{vmatrix} \]

### Step 4: Factor out \( \log r \)
We can factor out \( \log r \) from the first column:

\[ D = \log r \begin{vmatrix} p - r & p & 1 \\ q - r & q & 1 \\ r - 1 & r & 1 \end{vmatrix} \]

### Step 5: Evaluate the remaining determinant
Now, we need to evaluate the determinant:

\[ \begin{vmatrix} p - r & p & 1 \\ q - r & q & 1 \\ r - 1 & r & 1 \end{vmatrix} \]

Using the determinant formula, we can simplify this determinant. The result will yield:

\[ = (p - r)(q - r) - (p - r)(r - 1) - (q - r)(r - 1) \]

After simplifying, we can see that this determinant will ultimately yield zero because the rows are linearly dependent (as they represent terms of a G.P.).

### Final Result
Thus, the value of the determinant \( D \) is:

\[ D = 0 \]