Explanation

(a)
${\therefore a}_{1},{a_{2}}_{.},...,{a_{n}}_{.}$are in GP.
$\Rightarrow a_{n},a_{n} + 2,a_{n + 4},\ldots$ are also in GP.
Now, $\left( a_{n + 2} \right)^{2} = a_{n}.a_{n + 4}$
$\Rightarrow 2\log{\left( a_{n + 2} \right) = \log{a_{n} + \log a_{n + 4}}}$
Similarly, $2\log{\left( a_{n + 8} \right) = \log a_{n + 6} + \log a_{n + 10}}$
Now, $\mathrm{\Delta} = \left| \begin{matrix}
\log{a_{n}\ } & \log{a_{n + 2}\ } & \log{a_{n + 4}\ } \\
\log{a_{n + 6}\ } & \log{a_{n + 8}\ } & \log{a_{n + 10}\ } \\
\log{a_{n + 12}\ } & \log{a_{n + 14}\ } & \log{a_{n + 16}\ }
\end{matrix} \right|$
Applying $C_{2} \rightarrow 2C_{2} - C_{1} - C_{3}$
$\left| \begin{matrix}
\log{a_{n}\ } & 2\log{a_{n + 2} - \log{a_{n} - \log a_{n + 4}}\ } & \log{a_{n + 4}\ } \\
\log{a_{n + 6}\ } & {2log}{a_{n + 8} - \log{a_{n + 6 -}\log a_{n + 10}}\ } & \log{a_{n + 10}\ } \\
\log{a_{n + 12}\ } & 2\log{a_{n + 14} - \log{a_{n + 12} - \log a_{n + 16}}\ } & \log{a_{n + 16}\ }
\end{matrix} \right|$
=$\left| \begin{matrix}
\log{a_{n}\ } & 0 & \log{a_{n + 4}\ } \\
\log{a_{n + 6}\ } & 0 & \log{a_{n + 10}\ } \\
\log{a_{n + 12}\ } & 0 & \log{a_{n + 16}\ }
\end{matrix} \right| = 0$