$\left| {\begin{array}{ccc} 1 & 1+ac & 1+bc \\ 1 & 1+ad & 1+bd \\ 1 & 1+ae & 1+be \end{array}} \right| = $

Let $a, b, c$ be such that $b + c \ne 0$. If $\left| \begin{array}{ccc} a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1 \end{array} \right| + \left| \begin{array}{ccc} a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ (-1)^{n+2} \cdot a & (-1)^{n+1} \cdot b & (-1)^n \cdot c \end{array} \right| = 0$,then $n$ is equal to:

If $1, \omega, \omega^2$ are the cube roots of unity,then $\Delta = \begin{vmatrix} 1 & \omega^n & \omega^{2n} \\ \omega^n & \omega^{2n} & 1 \\ \omega^{2n} & 1 & \omega^n \end{vmatrix}$ is equal to

$\left| {\begin{array}{*{20}{c}}{x + 1}&{x + 2}&{x + 4}\\{x + 3}&{x + 5}&{x + 8}\\{x + 7}&{x + 10}&{x + 14}\end{array}} \right| = $

If $\left|\begin{array}{ccc}a^{2} & b c & c^{2}+a c \\ a^{2}+a b & b^{2} & c a \\ a b & b^{2}+b c & c^{2}\end{array}\right|=k a^{2} b^{2} c^{2}$, then $k=$

Let $A$ be a square matrix of order $3 \times 3$,then $|5A| = $ (in $|A|$)