If $$\left| {\matrix{ a & {{a^2}} & {1 + {a^3}} \cr b & {{b^2}} & {1 + {b^3}} \cr c & {{c^2}} & {1 + {c^3}} \cr } } \right| = 0$$ and the vectors
$$\overrightarrow A = \left( {1,a,{a^2}} \right),\,\,\overrightarrow B = \left( {1,b,{b^2}} \right),\,\,\overrightarrow C = \left( {1,c,{c^2}} \right),$$ are non-coplannar, then the product $$abc=$$ .......

If $$\overrightarrow A \overrightarrow {\,B} \overrightarrow {\,C} $$ are three non-coplannar vectors, then -
$${{\overrightarrow A .\overrightarrow B \times \overrightarrow C } \over {\overrightarrow C \times \overrightarrow A .\overrightarrow B }} + {{\overrightarrow B .\overrightarrow A \times \overrightarrow C } \over {\overrightarrow C .\overrightarrow A \times \overrightarrow B }} = $$ ................

If $$\overrightarrow A = \left( {1,1,1} \right),\,\,\overrightarrow C = \left( {0,1, - 1} \right)$$ are given vectors, then a vector $$B$$ satifying the equations $$\overrightarrow A \times \overrightarrow B = \overrightarrow {\,C} $$ and $$\overrightarrow A .\overrightarrow B = \overrightarrow {3\,} $$ ..........

$$A, B, C$$ and $$D,$$ are four points in a plane with position vectors $$a, b, c$$ and $$d$$ respectively such that $$$\left( {\overrightarrow a - \overrightarrow d } \right)\left( {\overrightarrow b - \overrightarrow c } \right) = \left( {\overrightarrow b - \overrightarrow d } \right)\left( {\overrightarrow c - \overrightarrow a } \right) = 0$$$

The point $$D,$$ then, is the ................ of the triangle $$ABC.$$