If the area of the triangle with vertices $(x, 0), (1, 1)$ and $(0, 2)$ is $4$ square units,then a value of $x$ is

Let $x, y, z > 0$ be the $2^{nd}, 3^{rd}, 4^{th}$ terms of a $G.P.$ respectively,and $\Delta = \begin{vmatrix} x^k & x^{k+1} & x^{k+2} \\ y^k & y^{k+1} & y^{k+2} \\ z^k & z^{k+1} & z^{k+2} \end{vmatrix} = (r-1)^2 \left(1 - \frac{1}{r^2}\right)$,where $r$ is the common ratio. Then $k = \dots$

If $A = \begin{bmatrix} 5 & 5\alpha & \alpha \\ 0 & \alpha & 5\alpha \\ 0 & 0 & 5 \end{bmatrix}$ and $\operatorname{det}(A^2) = 25$, then $|\alpha| = $

If $\left|\begin{array}{ccc}2 & 2k & 1 \\ 1 & k-1 & 1 \\ 2 & 1 & k+1\end{array}\right|=Ak^2+Bk+C$, then $A+B+C=$

If $A_n = \begin{bmatrix} 1-n & n \\ n & 1-n \end{bmatrix}$,then $|A_1| + |A_2| + \dots + |A_{2021}| = $

The number of values of $\theta \in (0, \pi)$ for which the system of linear equations $x + 3y + 7z = 0$,$-x + 4y + 7z = 0$,and $(\sin 3\theta)x + (\cos 2\theta)y + 2z = 0$ has a non-trivial solution is: