If $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ be three non-zero vectors satisfying the condition $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} = \overrightarrow{\mathrm{c}}$ and  $\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}} = \overrightarrow{\mathrm{a}}$,  then which of the following always hold(s) good?

Given the following information about the non-zero vectors $\overrightarrow{\mathrm{A}}, \overrightarrow{\mathrm{B}}$ and $\overrightarrow{\mathrm{C}}$

(a) $(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}) \times \overrightarrow{\mathrm{A}} = 0$

(b) $\overrightarrow{\mathrm{B}}.\overrightarrow{\mathrm{B}} = 4$

(c) $\overrightarrow{\mathrm{A}}.\overrightarrow{\mathrm{B}} = –6$

(d)$\overrightarrow{\mathrm{B}}.\overrightarrow{\mathrm{C}} = 6$

Which one of the following holds good?

Let $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ be non-zero vectors and $\left|\overrightarrow{a}\right|=1$ and $\overrightarrow{r}$ is a non-zero vector such that $\overrightarrow{r}\times \overrightarrow{a}=\overrightarrow{b}$ and $\overrightarrow{r}·\overrightarrow{c}=1$, then

A vector $\hat{d}$ is equally inclined to three vectors $\overrightarrow{a} =\hat{i}–\hat{j}\mathrm{+}\hat{k}, \overrightarrow{b}= 2\hat{i}\mathrm{+}\hat{\mathrm{j}}$ and $\overrightarrow{c} = 3\hat{j}–2\hat{k}$.  Let $\overrightarrow{x}, \overrightarrow{y}, \overrightarrow{z}$ be  three vectors in the plane of$\overrightarrow{a}, \overrightarrow{b} ; \overrightarrow{b}, \overrightarrow{c} ; \overrightarrow{c}, \overrightarrow{a}$  respectively, then