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

If the points with position vectors $60 \hat{i}+3 \hat{j}$,$40 \hat{i}-8 \hat{j}$,and $a \hat{i}-52 \hat{j}$ are collinear,then $a$ is equal to

If the system of equations $ax + y + z = 0$,$x + by + z = 0$ and $x + y + cz = 0$,where $a, b, c \neq 1$,has a non-trivial solution,then the value of $\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c}$ is

Let $S$ be the set of all values of $\theta \in [-\pi, \pi]$ for which the system of linear equations
$x + y + \sqrt{3} z = 0$
$-x + (\tan \theta) y + \sqrt{7} z = 0$
$x + y + (\tan \theta) z = 0$
has a non-trivial solution. Then $\frac{120}{\pi} \sum_{\theta \in S} \theta$ is equal to

Evaluate the determinant: $\left|\begin{array}{ccc}0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0\end{array}\right|$

$\left|\begin{array}{rr}\sin 35^{\circ} & -\cos 35^{\circ} \\ \sin 55^{\circ} & \cos 55^{\circ}\end{array}\right|=$ . . . . . .