$$\int\limits_{0}^{2}\left(\left|2 x^{2}-3 x\right|+\left[x-\frac{1}{2}\right]\right) \mathrm{d} x$$, where [t] is the greatest integer function, is equal to :

Consider a curve $$y=y(x)$$ in the first quadrant as shown in the figure. Let the area $$\mathrm{A}_{1}$$ is twice the area $$\mathrm{A}_{2}$$. Then the normal to the curve perpendicular to the line $$2 x-12 y=15$$ does NOT pass through the point.

The equations of the sides $$\mathrm{AB}, \mathrm{BC}$$ and CA of a triangle ABC are $$2 x+y=0, x+\mathrm{p} y=39$$ and $$x-y=3$$ respectively and $$\mathrm{P}(2,3)$$ is its circumcentre. Then which of the following is NOT true?

If the length of the perpendicular drawn from the point $$P(a, 4,2)$$, a $$>0$$ on the line $$\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}$$ is $$2 \sqrt{6}$$ units and $$Q\left(\alpha_{1}, \alpha_{2}, \alpha_{3}\right)$$ is the image of the point P in this line, then $$\mathrm{a}+\sum\limits_{i=1}^{3} \alpha_{i}$$ is equal to :