If u, v and w (all positive) are the `p^(th), q^(th) and r^(th)` terms of a GP, the determinant of the matrix `({:("In",u,pl),("In",v,ql),("In",w,rl):})`is

Options

A. 0
B. 1
C. (p - q) (q - r) (r - p)
D. In u `xx` In v `xx` In w

Explanation

To find the determinant of the matrix formed by the logarithms of the terms of a geometric progression (GP), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the terms of the GP**:
Let the first term of the GP be \( a \) and the common ratio be \( r \). Then the \( p^{th}, q^{th}, \) and \( r^{th} \) terms can be expressed as:
\[ u = ar^{p-1}, \quad v = ar^{q-1}, \quad w = ar^{r-1} \]

2. **Take logarithms of the terms**:
We take the natural logarithm of each term:
\[ \log u = \log(ar^{p-1}) = \log a + (p-1) \log r \]
\[ \log v = \log(ar^{q-1}) = \log a + (q-1) \log r \]
\[ \log w = \log(ar^{r-1}) = \log a + (r-1) \log r \]

3. **Set up the matrix**:
We can now form the matrix with these logarithmic values:
\[ \begin{pmatrix} \log u & p & 1 \\ \log v & q & 1 \\ \log w & r & 1 \end{pmatrix} \]

4. **Substituting the logarithmic values**:
Substitute the logarithmic expressions into the matrix:
\[ \begin{pmatrix} \log a + (p-1) \log r & p & 1 \\ \log a + (q-1) \log r & q & 1 \\ \log a + (r-1) \log r & r & 1 \end{pmatrix} \]

5. **Perform row operations**:
To simplify the determinant, we can perform row operations. We can subtract the first row from the second and third rows:
\[ R_2 \leftarrow R_2 - R_1 \]
\[ R_3 \leftarrow R_3 - R_1 \]
This gives us:
\[ \begin{pmatrix} \log a + (p-1) \log r & p & 1 \\ (q - p) + ((q-1) - (p-1)) \log r & 0 & 0 \\ (r - p) + ((r-1) - (p-1)) \log r & 0 & 0 \end{pmatrix} \]

6. **Calculate the determinant**:
The determinant of a matrix with two rows of zeros is zero. Therefore, the determinant of the matrix is:
\[ \text{Det} = 0 \]

### Final Result:
The determinant of the matrix is:
\[ \boxed{0} \]