If $${\mathop{\rm Re}\nolimits} \left( {{{z - 1} \over {2z + i}}} \right) = 1$$, where z = x + iy, then the point (x, y) lies on a :

A vector $$\overrightarrow a = \alpha \widehat i + 2\widehat j + \beta \widehat k\left( {\alpha ,\beta \in R} \right)$$ lies in the plane of the vectors, $$\overrightarrow b = \widehat i + \widehat j$$ and $$\overrightarrow c = \widehat i - \widehat j + 4\widehat k$$. If $$\overrightarrow a $$ bisects the angle between $$\overrightarrow b $$ and $$\overrightarrow c $$, then:

Let $$\alpha $$ be a root of the equation x² + x + 1 = 0 and the
matrix A = $${1 \over {\sqrt 3 }}\left[ {\matrix{ 1 & 1 & 1 \cr 1 & \alpha & {{\alpha ^2}} \cr 1 & {{\alpha ^2}} & {{\alpha ^4}} \cr } } \right]$$

then the matrix A³¹ is equal to

Let x^k + y^k = a^k, (a, k > 0 ) and $${{dy} \over {dx}} + {\left( {{y \over x}} \right)^{{1 \over 3}}} = 0$$, then k is: