An ordered pair $(\alpha , \beta )$ for which the system of linear equations

$\left( {1 + \alpha } \right)x + \beta y + z = 2$ ; $\alpha x + \left( {1 + \beta } \right)y + z = 3$ ; $\alpha x  + \beta y + 2z = 2$ has a unique solution, is

If $\left|\begin{array}{cc}x & 2 \\ 18 & x\end{array}\right|=\left|\begin{array}{cc}6 & 2 \\ 18 & 6\end{array}\right|,$ then $x$ is equal to

The set of all values of $\lambda$ for which the system of linear  $2{x_1} – 2{x_2} + {x_3} = \lambda {x_1}\;,\;2{x_1} – 3{x_2} + 2{x_3} = \lambda {x_2}\;\;,$$\;\; – {x_1} + 2{x_2} = \lambda {x_3}$ has a non-trivial solution

Let $S$ be the set of all $\lambda \in \mathrm{R}$ for which the system of linear equations

$2 x-y+2 z=2$

$x-2 y+\lambda z=-4$

$x+\lambda y+z=4$

has no solution. Then the set $S$

Let $\omega $ be a complex number such that  $2\omega + 1 = z$ where $z = \sqrt { – 3} $ . If $\left| {\begin{array}{*{20}{c}}1&1&1\\1&{ – {\omega ^2} – 1}&{{\omega ^2}}\\1&{{\omega ^2}}&{{\omega ^7}}\end{array}} \right| = 3k$ then $k$ is equal to :