Let $S=\left\{z \in \mathbb{C}: z^2+\sqrt{6} i z-3=0\right\}$. Then $\sum\limits_{z \in S} z^8$ is equal to :

The sum of all possible values of $\theta \in[0,2 \pi]$, for which the system of equations :

$$ \begin{aligned} & x \cos 3 \theta-8 y-12 z=0 \\ & x \cos 2 \theta+3 y+3 z=0 \\ & x+y+3 z=0 \end{aligned} $$

has a non-trivial solution, is equal to :

Let $\mathrm{A}=\left[\begin{array}{lll}1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1\end{array}\right]$ and $\mathrm{B}=\left[\mathrm{b}_{i j}\right], 1 \leq i, j \leq 3$. If $\mathrm{B}=\mathrm{A}^{99}-\mathrm{I}$, then the value of $\frac{\mathrm{b}_{31}-\mathrm{b}_{21}}{\mathrm{~b}_{32}}$ is :

The sum $1+\frac{1}{2}\left(1^2+2^2\right)+\frac{1}{3}\left(1^2+2^2+3^2\right)+\ldots$ upto 10 terms is equal to :