Let $P$ be a matrix of order $3 \times 3$ such that all the entries in $P$ are from the set $\{-1, 0, 1\}$. Then,the maximum possible value of the determinant of $P$ 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

If $\omega$ is an imaginary cube root of unity and $\left|\begin{array}{ccc}x+\omega^2 & \omega & 1 \\ \omega & \omega^2 & 1+x \\ 1 & x+\omega & \omega^2\end{array}\right|=0$,then one of the values of $x$ is

If $A = \begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix}$,then $\det(A)$ is equal to

If $p + q + r = 0$ and $a + b + c = 0$,then the value of the determinant $\left| \begin{array}{ccc} pa & qb & rc \\ qc & ra & pb \\ rb & pc & qa \end{array} \right|$ is

$A$ set of values of $\theta$ for which the system of equations $(\sin 3 \theta) x-y+z=0$,$(\cos 2 \theta) x+4 y+3 z=0, 2 x+7 y+7 z=0$ has non-trivial solutions,is