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 :

Consider points $$A, B, C$$ and $$D$$ with position

vectors $$7\widehat i - 4\widehat j + 7\widehat k,\widehat i - 6\widehat j + 10\widehat k, - \widehat i - 3\widehat j + 4\widehat k$$ and $$5\widehat i - \widehat j + 5\widehat k$$ respectively. Then $$ABCD$$ is a :

If $$\overrightarrow u \,,\overrightarrow v $$ and $$\overrightarrow w $$ are three non-coplanar vectors, then $$\,\left( {\overrightarrow u + \overrightarrow v - \overrightarrow w } \right).\left( {\overrightarrow u - \overrightarrow v } \right) \times \left( {\overrightarrow v - \overrightarrow w} \right)$$ equals :