If the point (2, $$\alpha $$, $$\beta $$) lies on the plane which passes through the points (3, 4, 2) and (7, 0, 6) and is perpendicular to the plane 2x – 5y = 15, then 2$$\alpha $$ – 3$$\beta $$ is equal to

Let S_n = 1 + q + q² + . . . . . + qⁿ and T_n = 1 + $$\left( {{{q + 1} \over 2}} \right) + {\left( {{{q + 1} \over 2}} \right)^2}$$ + . . . . . .+ $${\left( {{{q + 1} \over 2}} \right)^n}$$ where q is a real number and q $$ \ne $$ 1. If   ¹⁰¹C₁ + ¹⁰¹C₂ . S₁ + .... + ¹⁰¹C₁₀₁ . S₁₀₀ = $$\alpha $$T₁₀₀ then $$\alpha $$ is equal to

Let $$\sqrt 3 \widehat i + \widehat j,$$    $$\widehat i + \sqrt 3 \widehat j$$  and   $$\beta \widehat i + \left( {1 - \beta } \right)\widehat j$$ respectively be the position vectors of the points A, B and C with respect to the origin O. If the distance of C from the bisector of the acute angle between OA and OB is $${3 \over {\sqrt 2 }}$$, then the sum of all possible values of $$\beta $$ is :

All x satisfying the inequality (cot^–1 x)²– 7(cot^–1 x) + 10 > 0, lie in the interval :