If $\omega$ is a cube root of unity,then the root of the equation $\left| \begin{array}{ccc} x + 2 & \omega & \omega^2 \\ \omega & x + 1 + \omega^2 & 1 \\ \omega^2 & 1 & x + 1 + \omega \end{array} \right| = 0$ is:

If $\left| \begin{matrix} -6 & 1 & \lambda \\ 0 & 3 & 7 \\ -1 & 0 & 5 \end{matrix} \right| = 5948$,then $\lambda$ is:

Given that, $a \alpha^2+2 b \alpha+c \neq 0$ and that the system of equations
$\begin{aligned} & (a \alpha+b) x+a y+b z=0 \\ & (b \alpha+c) x+b y+c z=0 \\ & (a \alpha+b) y+(b \alpha+c) z=0\end{aligned}$
has a non-trivial solution, then $a, b$ and $c$ lie in

$\left| \begin{array}{ccc} 0 & p-q & p-r \\ q-p & 0 & q-r \\ r-p & r-q & 0 \end{array} \right| = $

$\left| \begin{array}{ccc} 2 \sin \frac{\pi}{3} & 1 & 0 \\ 1 & 2 \sin \frac{\pi}{3} & 1 \\ 0 & 1 & 2 \cos \frac{\pi}{6} \end{array} \right| = $ . . . . . .

The set of all values of $t \in R$,for which the matrix $\left[\begin{array}{ccc}e^t & e^{-t}(\sin t-2 \cos t) & e^{-t}(-2 \sin t-\cos t) \\e^t & e^{-t}(2 \sin t+\cos t) & e^{-t}(\sin t-2 \cos t) \\e^t & e^{-t} \cos t & e^{-t} \sin t \end{array}\right]$ is invertible.