Given that $a, b, c$ are $p^{\;\text{th}\;}, q^{\;\text{th}\;}$ and $r^{\;\text{th}\;}$ terms of a G.P. then, $A_{p}=AR^{p-1}=a, A_{q}=AR^{q-1}=b, A_{r}=AR^{r-1}=c$, where $A$ and $R$ are the $1^{\;\text{st}\;}$ term and common ratio of the geometric progression respectively. Consider LHS : Let $\Delta=\begin{vmatrix} \log a & p & 1 \\ \log b & q & 1 \\ \log c & r & 1 \end{vmatrix}$ $\Rightarrow \Delta=\begin{vmatrix} \log [AR^{p-1}] & p & 1 \\ \log [AR^{q-1}] & q & 1 \\ \log [AR^{r-1}] & r & 1 \end{vmatrix}$ $\left[\begin{array}{l} \because \log (mn)=\log m+\log n \\ \log (m)^n=n\log m \end{array}\right].$ $\therefore \Delta=\begin{vmatrix} \log A+(p-1)\log R & p & 1 \\ \log A+(q-1)\log R & q & 1 \\ \log A+(r-1)\log R & r & 1 \end{vmatrix}$By $C_1 \rightarrow C_1-(\log A)C_3$$\Rightarrow \Delta=\begin{vmatrix} (p-1)\log R & p & 1 \\ (q-1)\log R & q & 1 \\ (r-1)\log R & r & 1 \end{vmatrix}$Taking $\log R$ common from $C_1$$\Rightarrow \Delta=\log R\begin{vmatrix} p-1 & p & 1 \\ q-1 & q & 1 \\ r-1 & r & 1 \end{vmatrix}$By $C_1 \rightarrow C_1+C_3$$\Rightarrow \Delta=\log R\begin{vmatrix} p & p & 1 \\ q & q & 1 \\ r & r & 1 \end{vmatrix}$Since $C_1$ and $C_2$ are identical, $\therefore \Delta=0=$ RHS.