If $$\overrightarrow \alpha $$ = $$\left( {\lambda - 2} \right)\overrightarrow a + \overrightarrow b $$  and  $$\overrightarrow \beta = \left( {4\lambda - 2} \right)\overrightarrow a + 3\overrightarrow b $$ be two given vectors $$\overrightarrow a $$ and $$\overrightarrow b $$ are non-collinear. The value of $$\lambda $$ for which vectors $$\overrightarrow \alpha $$ and $$\overrightarrow \beta $$ are collinear, is -

Let $$z = {\left( {{{\sqrt 3 } \over 2} + {i \over 2}} \right)^5} + {\left( {{{\sqrt 3 } \over 2} - {i \over 2}} \right)^5}.$$ If R(z) and 1(z) respectively denote the real and imaginary parts of z, then :

Let A = $$\left[ {\matrix{ 2 & b & 1 \cr b & {{b^2} + 1} & b \cr 1 & b & 2 \cr } } \right]$$ where b > 0.

Then the minimum value of $${{\det \left( A \right)} \over b}$$ is -

The value of $$\lambda $$ such that sum of the squares of the roots of the quadratic equation, x² + (3 – $$\lambda $$)x + 2 = $$\lambda $$ has the least value is -