$\begin{vmatrix} 10! &11! &12! \\ 11! & 12! & 13!\\ 12! &13! & 14! \end{vmatrix}$
$\begin{vmatrix} 10! &11.10! &12.11.10! \\ 11! & 12.11! & 13.12.11!\\ 12! &13.12! & 14.13.12! \end{vmatrix}$
Taking 10! common from R1, 11! from R2, 12! from R3.
(10!)(11!)(12!) $\begin{vmatrix} 1 &11 &12.11 \\ 1 & 12 & 13.12\\ 1 &13 & 14.13 \end{vmatrix}$
R2→ R2 – R1
R3→ R3 – R2
(10!)(11!)(12!) $\begin{vmatrix} 1 &11 &12.11 \\ 0 & 1 & 24\\ 0 &1& 26 \end{vmatrix}$
R3→ R3 – R2
(10!)(11!)(12!)$\begin{vmatrix} 1 &11 &12.11 \\ 0 & 1 & 24\\ 0 &0& 2 \end{vmatrix}$
2(10!)(11!)(12!). Therefore, K=2.