If \(a_{1} a_{2} a_{3} \ldots a_{9}\) are in \(\mathrm{AP}\), then the value of \(\left|\begin{array}{lll}a_{1} & a_{2} & a_{3} \\ a_{4} & a_{5} & a_{6} \\ a_{7} & a_{8} & a_{9}\end{array}\right|\) is

\(a_{1}, a_{2}, a_{3}, \ldots, a_{9}\) are in AP.
\(\left|\begin{array}{lll}
a_{1} & a_{2} & a_{3} \\
a_{4} & a_{5} & a_{6} \\
a_{7} & a_{8} & a_{9}
\end{array}\right|\)
Apply \(R_{1} \rightarrow R_{1}+R_{2}-2 R_{2}\)
\(\left|\begin{array}{ccc}
a_{1}+a_{7}-2 a_{4} & a_{2}+a_{8}-2 a_{5} & a_{3}+a_{9}-2 a_{6} \\
a_{4} & a_{5} & a_{6} \\
a_{7} & a_{8} & a_{9}
\end{array}\right|\)
\(\left|\begin{array}{ccc}
0 & 0 & 0 \\
a_{4} & a_{5} & a_{6} \\
a_{7} & a_{8} & a_{9}
\end{array}\right|\)
\(\begin{aligned}
&{\left[\because a_{1}+a_{7}-2 a_{4}=a_{1}+a_{1}+6 d-2 a_{1}-6 d=0\right.} \\
&\text { Similarly } \left.a_{2}+a_{4}-2 a_{5}=0, a_{3}+a_{9}-2 a_{6}=0\right] \\
&=0 \\
&=\log _{e} 1=\log _{e}\left(\log _{e} e\right)
\end{aligned}\)