Let $$\overrightarrow a $$ = 2$$\widehat i$$ $$-$$ 3$$\widehat j$$ + 4$$\widehat k$$ and $$\overrightarrow b $$ = 7$$\widehat i$$ + $$\widehat j$$ $$-$$ 6$$\widehat k$$.

If $$\overrightarrow r $$ $$\times$$ $$\overrightarrow a $$ = $$\overrightarrow r $$ $$\times$$ $$\overrightarrow b $$, $$\overrightarrow r $$ . ($$\widehat i$$ + 2$$\widehat j$$ + $$\widehat k$$) = $$-$$3, then $$\overrightarrow r $$ . (2$$\widehat i$$ $$-$$ 3$$\widehat j$$ + $$\widehat k$$) is equal to :

Let $$\overrightarrow a $$ = $$\widehat i$$ + 2$$\widehat j$$ $$-$$ 3$$\widehat k$$ and $$\overrightarrow b = 2\widehat i$$ $$-$$ 3$$\widehat j$$ + 5$$\widehat k$$. If $$\overrightarrow r $$ $$\times$$ $$\overrightarrow a $$ = $$\overrightarrow b $$ $$\times$$ $$\overrightarrow r $$,

$$\overrightarrow r $$ . $$\left( {\alpha \widehat i + 2\widehat j + \widehat k} \right)$$ = 3 and $$\overrightarrow r \,.\,\left( {2\widehat i + 5\widehat j - \alpha \widehat k} \right)$$ = $$-$$1, $$\alpha$$ $$\in$$ R, then the

value of $$\alpha$$ + $${\left| {\overrightarrow r } \right|^2}$$ is equal to :

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 :