Let f be a real valued function, defined on R $$-$$ {$$-$$1, 1} and given by

f(x) = 3 log_e $$\left| {{{x - 1} \over {x + 1}}} \right| - {2 \over {x - 1}}$$.

Then in which of the following intervals, function f(x) is increasing?

Let $$\overrightarrow a $$ = $$\widehat i$$ + 2$$\widehat j$$ $$-$$ 3$$\widehat k$$ and $$\overrightarrow b = 2\widehat i$$ $$-$$ 3$$\widehat j$$ + 5$$\widehat k$$. If $$\overrightarrow r $$ $$\times$$ $$\overrightarrow a $$ = $$\overrightarrow b $$ $$\times$$ $$\overrightarrow r $$,

$$\overrightarrow r $$ . $$\left( {\alpha \widehat i + 2\widehat j + \widehat k} \right)$$ = 3 and $$\overrightarrow r \,.\,\left( {2\widehat i + 5\widehat j - \alpha \widehat k} \right)$$ = $$-$$1, $$\alpha$$ $$\in$$ R, then the

value of $$\alpha$$ + $${\left| {\overrightarrow r } \right|^2}$$ is equal to :

If the foot of the perpendicular from point (4, 3, 8) on the line $${L_1}:{{x - a} \over l} = {{y - 2} \over 3} = {{z - b} \over 4}$$, l $$\ne$$ 0 is (3, 5, 7), then the shortest distance between the line L₁ and line $${L_2}:{{x - 2} \over 3} = {{y - 4} \over 4} = {{z - 5} \over 5}$$ is equal to :

Consider a rectangle ABCD having 5, 7, 6, 9 points in the interior of the line segments AB, CD, BC, DA respectively. Let $$\alpha$$ be the number of triangles having these points from different sides as vertices and $$\beta$$ be the number of quadrilaterals having these points from different sides as vertices. Then ($$\beta$$ $$-$$ $$\alpha$$) is equal to :