$$P\left( {A \cup B} \right) = P\left( {A \cap B} \right)$$ if and only if the relation between $$P(A)$$ and $$P(B)$$ is .............

In a multiple-choice question there are four alternative answers, of which one or more are correct. A candidate will get marks in the question only if he ticks the correct answers. The candidate decides to tick the answers at random, If he is allowed upto three chances to answer the questions, find the probability that he will get marks in the questions.

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 -
$${{\overrightarrow A .\overrightarrow B \times \overrightarrow C } \over {\overrightarrow C \times \overrightarrow A .\overrightarrow B }} + {{\overrightarrow B .\overrightarrow A \times \overrightarrow C } \over {\overrightarrow C .\overrightarrow A \times \overrightarrow B }} = $$ ................