Let $$\overrightarrow \alpha = 3\widehat i + \widehat j$$ and $$\overrightarrow \beta = 2\widehat i - \widehat j + 3 \widehat k$$ . If $$\overrightarrow \beta = {\overrightarrow \beta _1} - \overrightarrow {{\beta _2}} $$, where $${\overrightarrow \beta _1}$$ is parallel to $$\overrightarrow \alpha $$ and $$\overrightarrow {{\beta _2}} $$ is perpendicular to $$\overrightarrow \alpha $$ , then $${\overrightarrow \beta _1} \times \overrightarrow {{\beta _2}} $$ is equal to

All the points in the set
$$S = \left\{ {{{\alpha + i} \over {\alpha - i}}:\alpha \in R} \right\}(i = \sqrt { - 1} )$$ lie on a :

If $$\left[ {\matrix{ 1 & 1 \cr 0 & 1 \cr } } \right]\left[ {\matrix{ 1 & 2 \cr 0 & 1 \cr } } \right]$$$$\left[ {\matrix{ 1 & 3 \cr 0 & 1 \cr } } \right]$$....$$\left[ {\matrix{ 1 & {n - 1} \cr 0 & 1 \cr } } \right] = \left[ {\matrix{ 1 & {78} \cr 0 & 1 \cr } } \right]$$,

then the inverse of $$\left[ {\matrix{ 1 & n \cr 0 & 1 \cr } } \right]$$ is

If the function ƒ : R – {1, –1} $$ \to $$ A defined by
ƒ(x) = $${{{x^2}} \over {1 - {x^2}}}$$ , is surjective, then A is equal to