Let $z_1, z_2 \in \mathbb{C}$ be the distinct solutions of the equation $z^2+4 z-(1+12 i)=0$.

Then $\left|z_1\right|^2+\left|z_2\right|^2$ is equal to :

If $f: \mathbf{N} \rightarrow \mathbf{Z}$ is defined by

$$ f(n)=\left|\begin{array}{ccc} n & -1 & -5 \\ -2 n^2 & 3(2 k+1) & 2 k+1 \\ -3 n^3 & 3 k(2 k+1) & 3 k(k+2)+1 \end{array}\right|, k \in N, $$

and $\sum\limits_{n=1}^k f(n)=98$, then $k$ is equal to :

Let M be a $3 \times 3$ matrix such that $\mathrm{M}\left(\begin{array}{l}1 \\ 0 \\ 0\end{array}\right)=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right), \mathrm{M}\left(\begin{array}{l}0 \\ 1 \\ 0\end{array}\right)=\left(\begin{array}{l}0 \\ 1 \\ 2\end{array}\right)$ and $\mathrm{M}\left(\begin{array}{l}0 \\ 0 \\ 1\end{array}\right)=\left(\begin{array}{c}-1 \\ 1 \\ 1\end{array}\right)$. If $\mathrm{M}\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{c}1 \\ 7 \\ 11\end{array}\right)$, then $x+y+z$ equals :

If the sum of the first 10 terms of the series $\frac{1}{1+1^4 \times 4}+\frac{2}{1+2^4 \times 4}+\frac{3}{1+3^4 \times 4}+\frac{4}{1+4^4 \times 4}+\ldots \ldots$. is $\frac{m}{n}, \operatorname{gcd}(m, n)=1$, then $m+n$ is equal to :