$a$ 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 $\sqrt{3}$)

Let the vectors $\vec{a}=(1+t) \hat{i}+(1-t) \hat{j}+\hat{k}$,$\vec{b}=(1-t) \hat{i}+(1+t) \hat{j}+2 \hat{k}$ and $\vec{c}=\hat{i}-t \hat{j}+\hat{k}$,$t \in R$ be such that for $\alpha, \beta, \gamma \in R$,$\alpha \vec{a}+\beta \vec{b}+\gamma \vec{c}=\vec{0} \Rightarrow \alpha=\beta=\gamma=0$. Then,the set of all values of $t$ is:

The volume of a tetrahedron whose vertices are $A \equiv (-1, 2, 3)$,$B \equiv (3, -2, 1)$,$C \equiv (2, 1, 3)$,and $D \equiv (-1, -2, 4)$ is

If for vectors $\bar{a}, \bar{b},$ and $\bar{c},$ $[\bar{a} \bar{b} \bar{c}] = 4,$ then $[\bar{a} \times \bar{b}, \bar{b} \times \bar{c}, \bar{c} \times \bar{a}] = \dots$

The volume of the tetrahedron having the edges $\hat{i}+2\hat{j}-\hat{k}$, $\hat{i}+\hat{j}+\hat{k}$, and $\hat{i}-\hat{j}+\lambda\hat{k}$ as coterminous edges is $\frac{2}{3}$ cubic units. Then $\lambda$ equals:

If $a(\alpha \times \beta)+b(\beta \times \gamma)+c(\gamma \times \alpha)=0$ and at least one of the scalars $a, b, c$ is non-zero,then the vectors $\alpha, \beta, \gamma$ are