If $${\overrightarrow a ,\overrightarrow b ,\overrightarrow c }$$ are non-coplanar vectors and $$\lambda $$ is a real number, then the vectors $${\overrightarrow a + 2\overrightarrow b + 3\overrightarrow c ,\,\,\lambda \overrightarrow b + 4\overrightarrow c }$$ and $$\left( {2\lambda - 1} \right)\overrightarrow c $$ are non coplanar for :

Let $$\overrightarrow u ,\overrightarrow v ,\overrightarrow w $$ be such that $$\left| {\overrightarrow u } \right| = 1,\,\,\,\left| {\overrightarrow v } \right|2,\,\,\,\left| {\overrightarrow w } \right|3.$$ If the projection $${\overrightarrow v }$$ along $${\overrightarrow u }$$ is equal to that of $${\overrightarrow w }$$ along $${\overrightarrow u }$$ and $${\overrightarrow v },$$ $${\overrightarrow w }$$ are perpendicular to each other then $$\left| {\overrightarrow u - \overrightarrow v + \overrightarrow w } \right|$$ equals :

A tetrahedron has vertices at $$O(0,0,0), A(1,2,1) B(2,1,3)$$ and $$C(-1,1,2).$$ Then the angle between the faces $$OAB$$ and $$ABC$$ will be :

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 vectors $$\left( {1,a,{a^2}} \right),\,\,$$

$$\left( {1,b,{b^2}} \right)$$ and $$\left( {1,c,{c^2}} \right)\,$$ are non-coplanar, then the product $$abc$$ equals :