$i \cdot (j \times k) + j \cdot (k \times i) + k \cdot (i \times j) = $

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:

$\vec{a}$ is a vector perpendicular to the plane containing non-zero vectors $\vec{b}$ and $\vec{c}$. If $\vec{a}, \vec{b}, \vec{c}$ are such that $|\vec{a}+\vec{b}+\vec{c}|=\sqrt{|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2}$, then $|(\vec{a} \times \vec{b}) \cdot \vec{c}|+|(\vec{a} \times \vec{b}) \times \vec{c}|=$

Let $a, b$ and $c$ be distinct non-negative numbers. If the vectors $a\hat{i} + a\hat{j} + c\hat{k}$,$\hat{i} + \hat{k}$ and $c\hat{i} + c\hat{j} + b\hat{k}$ are coplanar,then $c = \dots$

If the origin and the points $(1, 2, 3)$,$(2, 3, 4)$,and $(x, y, z)$ are coplanar,then

If $[\bar{a} \bar{b} \bar{c}] \neq 0$,then $\frac{[\bar{a}+\bar{b} \quad \bar{b}+\bar{c} \quad \bar{c}+\bar{a}]}{[\bar{b} \bar{c} \bar{a}]}=$