Let the line $$\,\,\,\,\,$$ $${{x - 2} \over 3} = {{y - 1} \over { - 5}} = {{z + 2} \over 2}$$ lie in the plane $$\,\,\,\,\,$$ $$x + 3y - \alpha z + \beta = 0.$$ Then $$\left( {\alpha ,\beta } \right)$$ equals

The projections of a vector on the three coordinate axis are $$6,-3,2$$ respectively. The direction cosines of the vector are :

If $$\overrightarrow u ,\overrightarrow v ,\overrightarrow w $$ are non-coplanar vectors and $$p,q$$ are real numbers, then the equality $$\left[ {3\overrightarrow u \,\,p\overrightarrow v \,\,p\overrightarrow w } \right] - \left[ {p\overrightarrow v \,\,\overrightarrow w \,\,q\overrightarrow u } \right] - \left[ {2\overrightarrow w \,\,q\overrightarrow v \,\,q\overrightarrow u } \right] = 0$$ holds for :

Let $$f\left( x \right) = {\left( {x + 1} \right)^2} - 1,x \ge - 1$$

Statement - 1 : The set $$\left\{ {x:f\left( x \right) = {f^{ - 1}}\left( x \right)} \right\} = \left\{ {0, - 1} \right\}$$.

Statement - 2 : $$f$$ is a bijection.