Let $$\overrightarrow a = - \widehat i - \widehat j + \widehat k,\overrightarrow a \,.\,\overrightarrow b = 1$$ and $$\overrightarrow a \times \overrightarrow b = \widehat i - \widehat j$$. Then $$\overrightarrow a - 6\overrightarrow b $$ is equal to :

If the four points, whose position vectors are $$3\widehat i - 4\widehat j + 2\widehat k,\widehat i + 2\widehat j - \widehat k, - 2\widehat i - \widehat j + 3\widehat k$$ and $$5\widehat i - 2\alpha \widehat j + 4\widehat k$$ are coplanar, then $$\alpha$$ is equal to :

Let $$A = \left[ {\matrix{ {{1 \over {\sqrt {10} }}} & {{3 \over {\sqrt {10} }}} \cr {{{ - 3} \over {\sqrt {10} }}} & {{1 \over {\sqrt {10} }}} \cr } } \right]$$ and $$B = \left[ {\matrix{ 1 & { - i} \cr 0 & 1 \cr } } \right]$$, where $$i = \sqrt { - 1} $$. If $$\mathrm{M=A^T B A}$$, then the inverse of the matrix $$\mathrm{AM^{2023}A^T}$$ is

$$\sum\limits_{k = 0}^6 {{}^{51 - k}{C_3}} $$ is equal to :