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:

If $(x_{1}, y_{1}), (x_{2}, y_{2})$ and $(x_{3}, y_{3})$ are the vertices of a triangle whose area is $k$ square units,then $\left|\begin{array}{ccc}x_{1} & y_{1} & 4 \\ x_{2} & y_{2} & 4 \\ x_{3} & y_{3} & 4\end{array}\right|^{2}$ is (in $k^{2}$)

Let $M=\left\{A=\begin{bmatrix} a & b \\ c & d \end{bmatrix} : a, b, c, d \in \{\pm 3, \pm 2, \pm 1, 0\}\right\}$. Define $f: M \rightarrow \mathbb{Z}$ as $f(A) = \det(A)$ for all $A \in M$,where $\mathbb{Z}$ is the set of all integers. Then the number of $A \in M$ such that $f(A) = 15$ is equal to $.....$

If $D = \left| \begin{array}{ccc} \frac{1}{z} & \frac{1}{z} & -\frac{(x+y)}{z^2} \\ -\frac{(y+z)}{x^2} & \frac{1}{x} & \frac{1}{x} \\ -\frac{y(y+z)}{x^2z} & \frac{x+2y+z}{xz} & -\frac{y(x+y)}{xz^2} \end{array} \right|$,then the incorrect statement is:

The area of a triangle with vertices $A(k, 1)$,$B(2, 4)$,and $C(1, 1)$ is $6$ sq. units. Find the value of $k$.

Find the values of $k$ if the area of the triangle is $4$ square units and the vertices are $(-2, 0), (0, 4), (0, k)$.