Let $$C$$ be the circle of minimum area touching the parabola $$y=6-x^2$$ and the lines $$y=\sqrt{3}|x|$$. Then, which one of the following points lies on the circle $$C$$ ?

If $$A(3,1,-1), B\left(\frac{5}{3}, \frac{7}{3}, \frac{1}{3}\right), C(2,2,1)$$ and $$D\left(\frac{10}{3}, \frac{2}{3}, \frac{-1}{3}\right)$$ are the vertices of a quadrilateral $$A B C D$$, then its area is

Let the relations $$R_1$$ and $$R_2$$ on the set $$X=\{1,2,3, \ldots, 20\}$$ be given by $$R_1=\{(x, y): 2 x-3 y=2\}$$ and $$R_2=\{(x, y):-5 x+4 y=0\}$$. If $$M$$ and $$N$$ be the minimum number of elements required to be added in $$R_1$$ and $$R_2$$, respectively, in order to make the relations symmetric, then $$M+N$$ equals

Let the area of the region enclosed by the curves $$y=3 x, 2 y=27-3 x$$ and $$y=3 x-x \sqrt{x}$$ be $$A$$. Then $$10 A$$ is equal to