If the area of a triangle is $3$ sq. units whose vertices are $A(1, 3)$,$B(0, 0)$,and $C(k, 0)$,then $k$ is equal to:

The sum of all possible values of $\theta \in [0, 2\pi]$, for which the system of equations : $x \cos 3\theta - 8y - 12z = 0, x \cos 2\theta + 3y + 3z = 0, x + y + 3z = 0$ has a non-trivial solution, is equal to :

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

If $A = \begin{bmatrix} \alpha^2 & 5 \\ 5 & -\alpha \end{bmatrix}$ and $\det(A^{10}) = 1024$,then $\alpha = $

Find the equation of the line joining $(1, 2)$ and $(3, 6)$ using determinants.

If $\left|\begin{array}{ccc}x & 4 & 6 \\ 2 & 3 & -9 \\ 5 & 6 & 1\end{array}\right|+\left|\begin{array}{ccc}5 & 6 & 1 \\ 6 & 4 & 5 \\ 2 & 3 & -9\end{array}\right|=\left|\begin{array}{ccc}2 & 3 & -9 \\ 1-2 x & -8 & -11 \\ 5 & 6 & 1\end{array}\right|$,then $x=$ . . . . . .