Let A be a $3 \times 3$ matrix such that $\mathrm{A}+\mathrm{A}^{\mathrm{T}}=\mathrm{O}$. If $\mathrm{A}\left[\begin{array}{c}1 \\ -1 \\ 0\end{array}\right]=\left[\begin{array}{l}3 \\ 3 \\ 2\end{array}\right], \mathrm{A}^2\left[\begin{array}{c}1 \\ -1 \\ 0\end{array}\right]=\left[\begin{array}{c}-3 \\ 19 \\ -24\end{array}\right]$ and $\operatorname{det}(\operatorname{adj}(2 \operatorname{adj}(\mathrm{~A}+\mathrm{I})))=(2)^\alpha \cdot(3)^\beta \cdot(11)^\gamma, \alpha, \beta, \gamma$ are non-negative integers, then $\alpha+\beta+\gamma$ is equal to $\_\_\_\_$

Let ABC be a triangle. Consider four points $\mathrm{p}_1, \mathrm{p}_2, \mathrm{p}_3, \mathrm{p}_4$ on the side AB , five points $p_5, p_6, p_7, p_8, p_9$ on the side $B C$, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC . None of these points is a vertex of the triangle ABC . Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, \ldots, p_{13}$, is $\_\_\_\_$

Let $\alpha=\frac{-1+i \sqrt{3}}{2}$ and $\beta=\frac{-1-i \sqrt{3}}{2}, i=\sqrt{-1}$. If

$$ (7-7 \alpha+9 \beta)^{20}+(9+7 \alpha-7 \beta)^{20}+(-7+9 \alpha+7 \beta)^{20}+(14+7 \alpha+7 \beta)^{20}=m^{10}, $$

then $m$ is $\_\_\_\_$

If $\int(\sin x)^{\frac{-11}{2}}(\cos x)^{\frac{-5}{2}} d x= -\frac{p_1}{q_1}(\cot x)^{\frac{9}{2}}-\frac{p_2}{q_2}(\cot x)^{\frac{5}{2}}-\frac{p_3}{q_3}(\cot x)^{\frac{1}{2}}+\frac{p_4}{q_4}(\cot x)^{\frac{-3}{2}}+\mathrm{C}$, where $p_i$ and $q_i$ are positive integers with $\operatorname{gcd}\left(p_i, q_i\right)=1$ for $i=1,2,3,4$ and C is the constant of integration, then $\frac{15 p_1 p_2 p_3 p_4}{q_1 q_2 q_3 q_4}$ is equal to $\_\_\_\_$