Let $a ,b ,c $ be such that $b + c \ne 0$  if

$\left| {\begin{array}{*{20}{c}}a&{a + 1}&{a – 1}\\{ – b}&{b + 1}&{b – 1}\\c&{c – 1}&{c + 1}\end{array}} \right| + \left| {\begin{array}{*{20}{c}}{a + 1}&{b + 1}&{c – 1}\\{a – 1}&{b – 1}&{c + 1}\\{{{\left( { – 1} \right)}^{n + 2}} \cdot a}&{{{\left( { – 1} \right)}^{n + 1}} \cdot b}&{{{\left( { – 1} \right)}^n} \cdot c}\end{array}} \right| = 0$ then $n$ equals to

If $n \ne 3k$ and 1, $\omega ,{\omega ^2}$ are the cube roots of unity, then $\Delta = \left| {\,\begin{array}{*{20}{c}}1&{{\omega ^n}}&{{\omega ^{2n}}}\\{{\omega ^{2n}}}&1&{{\omega ^n}}\\{{\omega ^n}}&{{\omega ^{2n}}}&1\end{array}\,} \right|$ has the value

If $|A|$  denotes the value of the determinant of the square matrix $A$ of order $3$ , then $ |-2A|=$

If $\left| {\,\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}\,} \right| = k(a + b + c)({a^2} + {b^2} + {c^2}$ $ – bc – ca – ab)$, then $k =$

If $p{\lambda ^4} + q{\lambda ^3} + r{\lambda ^2} + s\lambda + t = $ $\left| {\,\begin{array}{*{20}{c}}{{\lambda ^2} + 3\lambda }&{\lambda – 1}&{\lambda + 3}\\{\lambda + 1}&{2 – \lambda }&{\lambda – 4}\\{\lambda – 3}&{\lambda + 4}&{3\lambda }\end{array}\,} \right|,$ the value of $t$ is