Let p, q and r be real numbers (p $$ \ne $$ q, r $$ \ne $$ 0), such that the roots of the equation $${1 \over {x + p}} + {1 \over {x + q}} = {1 \over r}$$ are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to :

If the angle between the lines, $${x \over 2} = {y \over 2} = {z \over 1}$$

and $${{5 - x} \over { - 2}} = {{7y - 14} \over p} = {{z - 3} \over 4}\,\,$$ is $${\cos ^{ - 1}}\left( {{2 \over 3}} \right),$$ then p is equal to :

If the area of the region bounded by the curves, $$y = {x^2},y = {1 \over x}$$ and the lines y = 0 and x= t (t >1) is 1 sq. unit, then t is equal to :

If $$x = \sqrt {{2^{\cos e{c^{ - 1}}}}} $$ and $$y = \sqrt {{2^{se{c^{ - 1}}t}}} \,\,\left( {\left| t \right| \ge 1} \right),$$ then $${{dy} \over {dx}}$$ is equal to :