$$\text { Let } f(x)=\left\{\begin{array}{lr} -2, & -2 \leq x \leq 0 \\ x-2, & 0< x \leq 2 \end{array} \text { and } \mathrm{h}(x)=f(|x|)+|f(x)| \text {. Then } \int_\limits{-2}^2 \mathrm{~h}(x) \mathrm{d} x\right. \text { is equal to: }$$

One of the points of intersection of the curves $$y=1+3 x-2 x^2$$ and $$y=\frac{1}{x}$$ is $$\left(\frac{1}{2}, 2\right)$$. Let the area of the region enclosed by these curves be $$\frac{1}{24}(l \sqrt{5}+\mathrm{m})-\mathrm{n} \log _{\mathrm{e}}(1+\sqrt{5})$$, where $$l, \mathrm{~m}, \mathrm{n} \in \mathbf{N}$$. Then $$l+\mathrm{m}+\mathrm{n}$$ is equal to

Let $$\alpha \in(0, \infty)$$ and $$A=\left[\begin{array}{lll}1 & 2 & \alpha \\ 1 & 0 & 1 \\ 0 & 1 & 2\end{array}\right]$$. If $$\operatorname{det}\left(\operatorname{adj}\left(2 A-A^T\right) \cdot \operatorname{adj}\left(A-2 A^T\right)\right)=2^8$$, then $$(\operatorname{det}(A))^2$$ is equal to:

There are 5 points $$P_1, P_2, P_3, P_4, P_5$$ on the side $$A B$$, excluding $$A$$ and $$B$$, of a triangle $$A B C$$. Similarly there are 6 points $$\mathrm{P}_6, \mathrm{P}_7, \ldots, \mathrm{P}_{11}$$ on the side $$\mathrm{BC}$$ and 7 points $$\mathrm{P}_{12}, \mathrm{P}_{13}, \ldots, \mathrm{P}_{18}$$ on the side $$\mathrm{CA}$$ of the triangle. The number of triangles, that can be formed using the points $$\mathrm{P}_1, \mathrm{P}_2, \ldots, \mathrm{P}_{18}$$ as vertices, is: