If the given vectors $(-bc, b^2 + bc, c^2 + bc)$,$(a^2 + ac, -ac, c^2 + ac)$ and $(a^2 + ab, b^2 + ab, -ab)$ are coplanar,where none of $a, b$ and $c$ is zero,then:

If $\vec{a}$ and $\vec{b}$ are mutually perpendicular unit vectors and $\vec{r}$ is a vector such that $\vec{r} \cdot \vec{a} = 0$,$\vec{r} \cdot \vec{b} = 1$,and $[\vec{r} \, \vec{a} \, \vec{b}] = 1$,then $\vec{r} = \dots$

If $\bar{x}=\frac{\bar{b} \times \bar{c}}{[\bar{a} \bar{b} \bar{c}]}, \bar{y}=\frac{\bar{c} \times \bar{a}}{[\bar{a} \bar{b} \bar{c}]}$ and $\bar{z}=\frac{\bar{a} \times \bar{b}}{[\bar{a} \bar{b} \bar{c}]}$ where $\bar{a}, \bar{b}, \bar{c}$ are non-coplanar vectors,then the value of $\bar{x} \cdot(\bar{a}+\bar{b})+\bar{y} \cdot(\bar{b}+\bar{c})+\bar{z} \cdot(\bar{c}+\bar{a})$ is

If $\bar{a}, \bar{b}, \bar{c}$ are mutually perpendicular vectors such that $|\bar{a}| = a, |\bar{b}| = b, |\bar{c}| = c$,then $[\bar{a} \bar{b} \bar{c}] = ......$

If the vectors $\hat{\imath}+\hat{\jmath}+\hat{k}$,$\hat{\imath}-\hat{\jmath}+\hat{k}$ and $2\hat{\imath}+3\hat{\jmath}+m\hat{k}$ are coplanar,then $m=$

If $a, b$ and $c$ are non-coplanar vectors and the four points with position vectors $2a+3b-c$,$a-2b+3c$,$3a+4b-2c$ and $ka-6b+6c$ are coplanar,then $k=$