Evaluate `Delta` using the properties of Determinant
`Delta=|{:(loga,p,1),(logb,q,1),(logc,r,1):}| "where" a,b,c (gt 0) "are the " p^(th),q^(th)and r^(th)` terms of a G.P

Answer

To evaluate the determinant \( \Delta = \begin{vmatrix} \log a & p & 1 \\ \log b & q & 1 \\ \log c & r & 1 \end{vmatrix} \), where \( a, b, c > 0 \) are the \( p^{th}, q^{th}, \) and \( r^{th} \) terms of a geometric progression (G.P.), we can follow these steps:

### Step 1: Understand the terms in G.P.
In a G.P., the \( n^{th} \) term can be expressed as:
\[ T_n = a \cdot r^{n-1} \]
where \( a \) is the first term and \( r \) is the common ratio. Therefore, we can express:
- \( a = T_p = a \cdot r^{p-1} \)
- \( b = T_q = a \cdot r^{q-1} \)
- \( c = T_r = a \cdot r^{r-1} \)

### Step 2: Substitute the values into the determinant
Now substituting these values into the determinant:
\[ \Delta = \begin{vmatrix} \log(a \cdot r^{p-1}) & p & 1 \\ \log(a \cdot r^{q-1}) & q & 1 \\ \log(a \cdot r^{r-1}) & r & 1 \end{vmatrix} \]

### Step 3: Simplify the logarithmic terms
Using the property of logarithms \( \log(xy) = \log x + \log y \):
\[ \Delta = \begin{vmatrix} \log a + (p-1) \log r & p & 1 \\ \log a + (q-1) \log r & q & 1 \\ \log a + (r-1) \log r & r & 1 \end{vmatrix} \]

### Step 4: Apply row operations
Now, we can perform row operations to simplify the determinant. We can subtract the first row from the second and third rows:
- New Row 2: \( R_2 - R_1 \)
- New Row 3: \( R_3 - R_1 \)

This gives us:
\[ \Delta = \begin{vmatrix} \log a + (p-1) \log r & p & 1 \\ (q - p) & 1 & 0 \\ (r - p) & 1 & 0 \end{vmatrix} \]

### Step 5: Expand the determinant
Now, we can expand the determinant along the last column:
\[ \Delta = 1 \cdot \begin{vmatrix} \log a + (p-1) \log r & p \\ (q - p) & 1 \end{vmatrix} \]

Calculating this 2x2 determinant:
\[ = \left( \log a + (p-1) \log r \right) \cdot 1 - p \cdot (q - p) \]
\[ = \log a + (p-1) \log r - pq + p^2 \]

### Step 6: Analyze the result
Since \( a, b, c \) are terms of a G.P., the ratios between the terms are constant, leading to a linear dependence among the rows of the determinant. This implies that:
\[ \Delta = 0 \]

### Final Result
Thus, the value of the determinant \( \Delta \) is:
\[ \Delta = 0 \]