The value of […] is

The correct answer is Option (2) → \( 2(9!.10!.11!) \)

Since \( 10!=10\cdot 9! \), \( 11!=11\cdot 10\cdot 9! \), and so on, 9!, 10! and 11! come out as common factors of the rows R₁, R₂ and R₃:

    \( \displaystyle \Delta=9!\,10!\,11!\begin{vmatrix}1 & 10 & 110\\1 & 11 & 132\\1 & 12 & 156\end{vmatrix} \)

Applying the row operations \( R_3\to R_3-R_2 \) and \( R_2\to R_2-R_1 \):

    \( \displaystyle \Delta=9!\,10!\,11!\begin{vmatrix}1 & 10 & 110\\0 & 1 & 22\\0 & 1 & 24\end{vmatrix} \)

Expanding along C₁:

    \( \displaystyle \Delta=9!\,10!\,11!\times(1\times 24-22\times 1) \)

    \( \displaystyle =2(9!\,10!\,11!) \)