For any three vectors $${\overrightarrow a ,\,\overrightarrow b ,}$$ and $${\overrightarrow c ,}$$
$$\left( {\overrightarrow a - \overrightarrow b } \right)\,.\,\left( {\overrightarrow b - \overrightarrow c } \right)\, \times \,\left( {\overrightarrow c - \overrightarrow a } \right)\, = \,2\overrightarrow {a\,} .\,\overrightarrow {b\,} \times \,\overrightarrow c .$$

The points with position vectors $$a+b,$$ $$a-b,$$ and $$a+kb$$ are collinear for all real values of $$k.$$

If $$X.A=0, X.B=0, X.C=0$$ for some non-zero vector $$X,$$ then $$\left[ {A\,B\,C} \right] = 0$$

Let $$\overrightarrow A ,\overrightarrow B $$ and $${\overrightarrow C }$$ be unit vectors suppose that $$\overrightarrow A .\overrightarrow B = \overrightarrow A .\overrightarrow C = 0,$$ and thatthe angle between $${\overrightarrow B }$$ and $${\overrightarrow C }$$ is $$\pi /6.$$ Then $$\overrightarrow A = \pm 2\left( {\overrightarrow B \times \overrightarrow C } \right).$$