Let a vector $$\alpha \widehat i + \beta \widehat j$$ be obtained by rotating the vector $$\sqrt 3 \widehat i + \widehat j$$ by an angle 45$$^\circ$$ about the origin in counterclockwise direction in the first quadrant. Then the area of triangle having vertices ($$\alpha$$, $$\beta$$), (0, $$\beta$$) and (0, 0) is equal to :

If vectors $$\overrightarrow {{a_1}} = x\widehat i - \widehat j + \widehat k$$ and $$\overrightarrow {{a_2}} = \widehat i + y\widehat j + z\widehat k$$ are collinear, then a possible unit vector parallel to the vector $$x\widehat i + y\widehat j + z\widehat k$$ is :

If $$\overrightarrow a $$ and $$\overrightarrow b $$ are perpendicular, then
$$\overrightarrow a \times \left( {\overrightarrow a \times \left( {\overrightarrow a \times \left( {\overrightarrow a \times \overrightarrow b } \right)} \right)} \right)$$ is equal to :

If the volume of a parallelopiped, whose
coterminus edges are given by the
vectors $$\overrightarrow a = \widehat i + \widehat j + n\widehat k$$,
$$\overrightarrow b = 2\widehat i + 4\widehat j - n\widehat k$$ and
$$\overrightarrow c = \widehat i + n\widehat j + 3\widehat k$$ ($$n \ge 0$$), is 158 cu. units, then :