Let $$\left(5, \frac{a}{4}\right)$$ be the circumcenter of a triangle with vertices $$\mathrm{A}(a,-2), \mathrm{B}(a, 6)$$ and $$C\left(\frac{a}{4},-2\right)$$. Let $$\alpha$$ denote the circumradius, $$\beta$$ denote the area and $$\gamma$$ denote the perimeter of the triangle. Then $$\alpha+\beta+\gamma$$ is

In a triangle ABC, if $$\cos \mathrm{A}+2 \cos \mathrm{B}+\cos C=2$$ and the lengths of the sides opposite to the angles A and C are 3 and 7 respectively, then $$\mathrm{\cos A-\cos C}$$ is equal to

For a triangle $$ABC$$, the value of $$\cos 2A + \cos 2B + \cos 2C$$ is least. If its inradius is 3 and incentre is M, then which of the following is NOT correct?

A straight line cuts off the intercepts $$\mathrm{OA}=\mathrm{a}$$ and $$\mathrm{OB}=\mathrm{b}$$ on the positive directions of $$x$$-axis and $$y$$ axis respectively. If the perpendicular from origin $$O$$ to this line makes an angle of $$\frac{\pi}{6}$$ with positive direction of $$y$$-axis and the area of $$\triangle \mathrm{OAB}$$ is $$\frac{98}{3} \sqrt{3}$$, then $$\mathrm{a}^{2}-\mathrm{b}^{2}$$ is equal to :