If the vectors $\vec{a}=\hat{i}+\hat{j}+\hat{k}$,$\vec{b}=\hat{i}-\hat{j}+2\hat{k}$ and $\vec{c}=x\hat{i}+(x-2)\hat{j}-\hat{k}$ are coplanar,then $x=$

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:

If a vector $\alpha$ lies in the plane of $\beta$ and $\gamma$,then which of the following is correct?

If $\bar{a}+\bar{b}, \bar{b}+\bar{c}$ and $\bar{c}+\bar{a}$ are coterminous edges of a parallelepiped,then its volume is $ . . . . . . $

Let the volume of tetrahedron $ABCD$ be $81$ cubic units and $G_1, G_2, G_3$ be the centroids of the triangular faces $ABC, ABD,$ and $ACD$ respectively. Then the volume of tetrahedron $AG_1G_2G_3$ is (in cubic units):

Let $\bar{a}=\lambda \bar{i}+3 \bar{j}+4 \bar{k}$,$\bar{b}=3 \bar{i}-\bar{j}+\lambda \bar{k}$ and $\bar{c}=\lambda \bar{i}+\bar{j}-3 \bar{k}$ be three vectors for some integer $\lambda$. If the volume of the parallelepiped with $\bar{a}, \bar{b}, \bar{c}$ as coterminus edges is $61$ cubic units,then the number of possible values of $\lambda$ is