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} = $

Let $\alpha$ and $\beta$ be the roots of the equation $x^2 + x + 1 = 0.$ Then for $y \ne 0$ in $\mathbb{R},$ the determinant $\left| \begin{array}{ccc} y + 1 & \alpha & \beta \\ \alpha & y + \beta & 1 \\ \beta & 1 & y + \alpha \end{array} \right|$ is equal to

The area of the triangle formed by the points $(a, b + c)$,$(b, c + a)$,and $(c, a + b)$ is:

If points $(5, 5)$,$(10, k)$ and $(-5, 1)$ are collinear,then $k =$

If $\omega$ is a cube root of unity and $\Delta = \begin{vmatrix} 1 & 2\omega \\ \omega & \omega^2 \end{vmatrix}$,then $\Delta^2$ is equal to

If $a, b, c$ are sides of $\triangle ABC$ and $\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0$,then $\sin^2 A + \sin^2 B + \sin^2 C = $ . . . . . . .