left|begin{array}{lll}125 & 5 & 25 343 & 7 & | vedclass

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(D) Let $\Delta = \left|\begin{array}{lll}125 & 5 & 25 \ 343 & 7 & 49 \ 729 & 9 & 81\end{array}ight|$.Taking $5$ common from $R_1$,$7$ common from

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$7$

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(D) Let $\Delta = \left|\begin{array}{lll}125 & 5 & 25 \ 343 & 7 & 49 \ 729 & 9 & 81\end{array}ight|$.Taking $5$ common from $R_1$,$7$ common from $R_2$,and $9$ common from $R_3$:$\Delta = (5 \cdot 7 \cdot 9) \left|\begin{array}{lll}25 & 1 & 5 \ 49 & 1 & 7 \ 81 & 1 & 9\end{array}ight|$.Applying $R_2 ightarrow R_2 - R_1$ and $R_3 ightarrow R_3 - R_1$:$\Delta = 315 \left|\begin{array}{lll}25 & 1 & 5 \ 24 & 0 & 2 \ 56 & 0 & 4\end{array}ight|$.Expanding along the second column:$\Delta = 315 \cdot (-1) \cdot \left|\begin{array}{ll}24 & 2 \ 56 & 4\end{array}ight|$.$\Delta = -315 \cdot (24 \cdot 4 - 56 \cdot 2) = -315 \cdot (96 - 112) = -315 \cdot (-16) = 5040$.Since $7! = 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 5040$,the result is $7!$.

$\left|\begin{array}{lll}125 & 5 & 25 \\ 343 & 7 & 49 \\ 729 & 9 & 81\end{array}\right|=$ (in $!$)

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