Consider $\overrightarrow{\mathrm{r}}, \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ are non-zero vectors such that $\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{a}}=0$, $|\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{r}}||\overrightarrow{\mathrm{b}}|,|\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{c}}|=|\overrightarrow{\mathrm{r}}||\overrightarrow{\mathrm{c}}|$, then $[\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}]$ is
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Solution
Here, it is given in problem $|\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{r}}||\overrightarrow{\mathrm{b}}|$
So, it is clear that angle between $\overrightarrow{\mathrm{r}}$ and $\overrightarrow{\mathrm{b}}$ is $\frac{\pi}{2}$ $\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{a}}=0 \Rightarrow \overrightarrow{\mathrm{r}}$ is perpendicular to $\overrightarrow{\mathrm{a}}$. $|\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{c}}|=|\overrightarrow{\mathrm{r}}| \cdot|\overrightarrow{\mathrm{c}}| \Rightarrow \overrightarrow{\mathrm{r}}$ is perpendicular to $\overrightarrow{\mathrm{c}}$. Thus, $\overrightarrow{\mathrm{r}}$ is perpendicular to $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$.
Hence, $\vec{a}, \vec{b}$ and $\vec{c}$ are coplanar.
$\Rightarrow[\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}]=0$