Given that $\mathrm{a},\mathrm{b},\mathrm{c}$ are and terms of a G.P. then,
${\mathrm{A}}_{\mathrm{p}}={\mathrm{AR}}^{\mathrm{p}-1}=\mathrm{a},{\mathrm{A}}_{\mathrm{q}}={\mathrm{AR}}^{\mathrm{q}-1}=\mathrm{b},{\mathrm{A}}_{\mathrm{r}}={\mathrm{AR}}^{\mathrm{r}-1}=\mathrm{c}$,
where $\mathrm{A}$ and $\mathrm{R}$ are the term and common ratio of the geometric progression respectively.
Consider LHS : Let $∆=\left|\begin{array}{ccc}\log \mathrm{a}& \mathrm{p}& 1\\ \log \mathrm{b}& \mathrm{q}& 1\\ \log \mathrm{c}& \mathrm{r}& 1\end{array}\right|$
$\Rightarrow ∆=\left|\begin{array}{ccc}\log \left[{\mathrm{AR}}^{\mathrm{p}-1}\right]& \mathrm{p}& 1\\ \log \left[{\mathrm{AR}}^{\mathrm{q}-1}\right]& \mathrm{q}& 1\\ \log \left[{\mathrm{AR}}^{\mathrm{r}-1}\right]& \mathrm{r}& 1\end{array}\right|$ $[\begin{array}{c}\because \log \left(\mathit{m}\mathit{n}\right)=\log \mathit{m}+\log \mathit{n}\\ \log {\left(\mathit{m}\right)}^{\mathit{n}}=\mathit{n}\log \mathit{m}\end{array}.$
$\therefore ∆=\left|\begin{array}{ccc}\log \mathrm{A}+\left(\mathrm{p}-1\right)\log \mathrm{R}& \mathrm{p}& 1\\ \log \mathrm{A}+\left(\mathrm{q}-1\right)\log \mathrm{R}& \mathrm{q}& 1\\ \log \mathrm{A}+\left(\mathrm{r}-1\right)\log \mathrm{R}& \mathrm{r}& 1\end{array}\right|$
By ${\mathrm{C}}_1\to {\mathrm{C}}_1-\left(\log \mathrm{A}\right){\mathrm{C}}_3$
$\Rightarrow ∆=\left|\begin{array}{ccc}\left(\mathrm{p}-1\right)\log \mathrm{R}& \mathrm{p}& 1\\ \left(\mathrm{q}-1\right)\log \mathrm{R}& \mathrm{q}& 1\\ \left(\mathrm{r}-1\right)\log \mathrm{R}& \mathrm{r}& 1\end{array}\right|$
Taking $\log \mathrm{R}$ common from ${\mathrm{C}}_1$
$\Rightarrow ∆=\log \mathit{R}\left|\begin{array}{ccc}\mathit{p}-1& \mathit{p}& 1\\ \mathit{q}-1& \mathit{q}& 1\\ \mathit{r}-1& \mathit{r}& 1\end{array}\right|$
By ${\mathrm{C}}_1\to {\mathrm{C}}_1+{\mathrm{C}}_3$
$\Rightarrow ∆=\log \mathit{R}\left|\begin{array}{ccc}\mathit{p}& \mathit{p}& 1\\ \mathit{q}& \mathit{q}& 1\\ \mathit{r}& \mathit{r}& 1\end{array}\right|$
Since ${\mathrm{C}}_1$ and ${\mathrm{C}}_2$ are identical, $\therefore ∆=0=$ RHS.