Let $f(x)=\int \frac{7 x^{10}+9 x^8}{\left(1+x^2+2 x^9\right)^2} d x, x>0, \lim\limits_{x \rightarrow 0} f(x)=0$ and $f(1)=\frac{1}{4}$.

If $\mathrm{A}=\left[\begin{array}{ccc}0 & 0 & 1 \\ \frac{1}{4} & f^{\prime}(1) & 1 \\ \alpha^2 & 4 & 1\end{array}\right]$ and $\mathrm{B}=\operatorname{adj}(\operatorname{adj} \mathrm{A})$ be such that $|\mathrm{B}|=81$, then $\alpha^2$ is equal to

The system of linear equations

$$ \begin{aligned} & x+y+z=6 \\ & 2 x+5 y+a z=36 \\ & x+2 y+3 z=b \end{aligned} $$

has :

Among the statements :

I: If $\left|\begin{array}{ccc}1 & \cos \alpha & \cos \beta \\ \cos \alpha & 1 & \cos \gamma \\ \cos \beta & \cos \gamma & 1\end{array}\right|=\left|\begin{array}{ccc}0 & \cos \alpha & \cos \beta \\ \cos \alpha & 0 & \cos \gamma \\ \cos \beta & \cos \gamma & 0\end{array}\right|$, then $\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=\frac{3}{2}$, and

II: If $\left|\begin{array}{ccc}x^2+x & x+1 & x-2 \\ 2 x^2+3 x-1 & 3 x & 3 x-3 \\ x^2+2 x+3 & 2 x-1 & 2 x-1\end{array}\right|=\mathrm{p} x+\mathrm{q}$, then $\mathrm{p}^2=196 \mathrm{q}^2$,

Let n be the number obtained on rolling a fair die. If the probability that the system

$$ \begin{aligned} & x-\mathrm{n} y+z=6 \\ & x+(\mathrm{n}-2) y+(\mathrm{n}+1) z=8 \\ & \quad(\mathrm{n}-1) y+z=1 \end{aligned} $$

has a unique solution is $\frac{k}{6}$, then the sum of $k$ and all possible values of $n$ is :