Given $a_1, a_2, a_3,..., a_9$ are in $G.P.,$ hence, by the definition of $G.P.,$ we have $\frac{a_2}{a_1}=\frac{a_3}{a_2}=\frac{a_4}{a_3}=...\frac{a_9}{a_8}=r,$ where $r$ is the common ratio of the $G.P.$

Now, let $△=\left|\begin{array}{lll}\log a_1& \log a_2& \log a_3\\ \log a_4& \log a_5& \log a_6\\ \log a_7& \log a_8& \log a_9\end{array}\right|$

Applying $C_2\to C_2-C_1, C_3\to C_3-C_2,$ we get

$△=\left|\begin{array}{lll}\log a_1& \log a_2-\log a_1& \log a_3-\log a_2\\ \log a_4& \log a_5-\log a_4& \log a_6-\log a_5\\ \log a_7& \log a_8-\log a_7& \log a_9-\log a_8\end{array}\right|$

Using $\log m-\log n=\log \left(\frac{m}{n}\right),$ we get

$\Rightarrow △=\left|\begin{array}{lll}\log a_1& \log \left(\frac{a_2}{a_1}\right)& \log \left(\frac{a_3}{a_2}\right)\\ \log a_4& \log \left(\frac{a_5}{a_4}\right)& \log \left(\frac{a_6}{a_5}\right)\\ \log a_7& \log \left(\frac{a_8}{a_7}\right)& \log \left(\frac{a_9}{a_8}\right)\end{array}\right|$

Using the relation defined above, we get

$\Rightarrow △=\left|\begin{array}{lll}\log a_1& \log \left(r\right)& \log \left(r\right)\\ \log a_4& \log \left(r\right)& \log \left(r\right)\\ \log a_7& \log \left(r\right)& \log \left(r\right)\end{array}\right|$

Thus, the determinant has $C_2=C_3,$ and we know that if two rows/columns of a determinant are same then its value is zero.

So, $△=\left|\begin{array}{lll}\log a_1& \log a_2& \log a_3\\ \log a_4& \log a_5& \log a_6\\ \log a_7& \log a_8& \log a_9\end{array}\right|=0.$