If z and $$\omega$$ are two complex numbers such that $$\left| {z\omega } \right| = 1$$ and $$\arg (z) - \arg (\omega ) = {{3\pi } \over 2}$$, then $$\arg \left( {{{1 - 2\overline z \omega } \over {1 + 3\overline z \omega }}} \right)$$ is :

(Here arg(z) denotes the principal argument of complex number z)

If in a triangle ABC, AB = 5 units, $$\angle B = {\cos ^{ - 1}}\left( {{3 \over 5}} \right)$$ and radius of circumcircle of $$\Delta$$ABC is 5 units, then the area (in sq. units) of $$\Delta$$ABC is :

Let [ x ] denote the greatest integer $$\le$$ x, where x $$\in$$ R. If the domain of the real valued function $$f(x) = \sqrt {{{\left| {[x]} \right| - 2} \over {\left| {[x]} \right| - 3}}} $$ is ($$-$$ $$\infty$$, a) $$]\cup$$ [b, c) $$\cup$$ [4, $$\infty$$), a < b < c, then the value of a + b + c is :

Let y = y(x) be the solution of the differential equation $$x\tan \left( {{y \over x}} \right)dy = \left( {y\tan \left( {{y \over x}} \right) - x} \right)dx$$, $$ - 1 \le x \le 1$$, $$y\left( {{1 \over 2}} \right) = {\pi \over 6}$$. Then the area of the region bounded by the curves x = 0, $$x = {1 \over {\sqrt 2 }}$$ and y = y(x) in the upper half plane is :