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 :

Let $$\overrightarrow u = \widehat i + \widehat j,\,\overrightarrow v = \widehat i - \widehat j$$ and $$\overrightarrow w = \widehat i + 2\widehat j + 3\widehat k\,\,.$$ If $$\widehat n$$ is a unit vector such that $$\overrightarrow u .\widehat n = 0$$ and $$\overrightarrow v .\widehat n = 0\,\,,$$ then $$\left| {\overrightarrow w .\widehat n} \right|$$ is equal to :

The vectors $$\overrightarrow {AB} = 3\widehat i + 4\widehat k\,\,\& \,\,\overrightarrow {AC} = 5\widehat i - 2\widehat j + 4\widehat k$$ are the sides of triangle $$ABC.$$ The length of the median through $$A$$ is :