If the vectors $$\overrightarrow a ,\overrightarrow b $$ and $$\overrightarrow c $$ form the sides $$BC,$$ $$CA$$ and $$AB$$ respectively of a triangle $$ABC,$$ then

Let the vectors $$\overrightarrow a ,\overrightarrow b ,\overrightarrow c $$ and $$\overrightarrow d $$ be such that
$$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \left( {\overrightarrow c \times \overrightarrow d } \right) = 0.$$ Let $${P_1}$$ and $${P_2}$$ be planes determined
by the pairs of vectors $$\overrightarrow a .\overrightarrow b $$ and $$\overrightarrow c .\overrightarrow d $$ respectively. Then the angle between $${P_1}$$ and $${P_2}$$ is

If $$\overrightarrow a \,,\,\overrightarrow b $$ and $$\overrightarrow c $$ are unit coplanar vectors, then the scalar triple product $$\left[ {2\overrightarrow a - \overrightarrow b ,2\overrightarrow b - \overrightarrow c ,2\overrightarrow c - \overrightarrow a } \right] = $$

If the normal to the curve $$y = f\left( x \right)$$ and the point $$(3, 4)$$ makes an angle $${{{3\pi } \over 4}}$$ with the positive $$x$$-axis, then $$f'\left( 3 \right) = $$