If $\left| \begin{array}{ccc} \cos 2x & \sin^2 x & \cos 4x \\ \sin^2 x & \cos 2x & \cos^2 x \\ \cos 4x & \cos^2 x & \cos 2x \end{array} \right| = a_0 + a_1 \sin x + a_2 \sin^2 x + \dots$,then $a_0$ is equal to:

Find the area of the triangle with vertices at the points $(-2, -3), (3, 2), (-1, -8)$.

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

The area of the triangle whose vertices are $(3,5), (2,2)$ and $(k, 2)$ is $3$ sq. unit. Then,the value of $k$ is . . . . . . .

If $x, y, z$ are in arithmetic progression with common difference $d$,$x \neq 3d$,and the determinant of the matrix $\begin{bmatrix} 3 & 4\sqrt{2} & x \\ 4 & 5\sqrt{2} & y \\ 5 & k & z \end{bmatrix}$ is zero,then the value of $k^2$ is ..... .

For positive numbers $x, y$ and $z$,the numerical value of the determinant $\left| \begin{array}{ccc} 1 & \log_x y & \log_x z \\ \log_y x & 1 & \log_y z \\ \log_z x & \log_z y & 1 \end{array} \right|$ is