$a$ is perpendicular to both $b$ and $c$. The angle between $b$ and $c$ is $\frac{2 \pi}{3}$. If $|a|=2$, $|b|=3$, and $|c|=4$, then $c \cdot (a \times b)$ is equal to (in $\sqrt{3}$)

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

For any non-zero vectors $\bar{a}$ and $\bar{b}$,$\left[\begin{array}{lll}\bar{b} & \bar{a} \times \bar{b} & \bar{a}\end{array}\right]=$

If $x, y$ and $z$ are non-zero real numbers and $\vec{a}=x \hat{i}+2 \hat{j}, \vec{b}=y \hat{j}+3 \hat{k}$ and $\vec{c}=x \hat{i}+y \hat{j}+z \hat{k}$ are such that $\vec{a} \times \vec{b}=z \hat{i}-3 \hat{j}+xy \hat{k}$ is not given,but $\vec{a} \times \vec{b}=6 \hat{i}-3 \hat{j}+\hat{k}$ is given as $z \hat{i}-3 \hat{j}+\hat{k}$,then the scalar triple product $[\vec{a} \vec{b} \vec{c}]$ is equal to:

If $\vec{a}, \vec{b}, \vec{c}$ are any three non-zero non-coplanar vectors and vectors $\vec{p} = \frac{\vec{b} \times \vec{c}}{[\vec{a} \vec{b} \vec{c}]}, \vec{q} = \frac{\vec{c} \times \vec{a}}{[\vec{a} \vec{b} \vec{c}]}, \vec{r} = \frac{\vec{a} \times \vec{b}}{[\vec{a} \vec{b} \vec{c}]}$,then $[\vec{p} \vec{q} \vec{r}] = ...$