The real part of the complex number $${{{{(1 + 2i)}^8}\,.\,{{(1 - 2i)}^2}} \over {(3 + 2i)\,.\,\overline {(4 - 6i)} }}$$ is equal to :

Let S be the set of all integral values of $$\alpha$$ for which the sum of squares of two real roots of the quadratic equation $$3{x^2} + (\alpha - 6)x + (\alpha + 3) = 0$$ is minimum. Then S :

Let $$A = \left[ {\matrix{ 1 & { - 2} & \alpha \cr \alpha & 2 & { - 1} \cr } } \right]$$ and $$B = \left[ {\matrix{ 2 & \alpha \cr { - 1} & 2 \cr 4 & { - 5} \cr } } \right],\,\alpha \in C$$. Then the absolute value of the sum of all values of $$\alpha$$ for which det(AB) = 0 is :

For two positive real numbers a and b such that $${1 \over {{a^2}}} + {1 \over {{b^3}}} = 4$$, then minimum value of the constant term in the expansion of $${\left( {a{x^{{1 \over 8}}} + b{x^{ - {1 \over {12}}}}} \right)^{10}}$$ is :