$\bar{a} = \hat{i} + \hat{j} + \hat{k}$,$\bar{b} = \hat{i} - \hat{j} + 2\hat{k}$ and $\bar{c} = x\hat{i} + (x-1)\hat{j} - \hat{k}$. If the vector $\bar{c}$ lies in the plane of $\bar{a}$ and $\bar{b}$,then $x=$

If $\alpha = 2i + 3j - k$,$\beta = -i + 2j - 4k$,and $\gamma = i + j + k$,then $(\alpha \times \beta) \cdot (\alpha \times \gamma)$ is equal to

Let $\overrightarrow{a}=\hat{i}-2 \hat{j}+3 \hat{k}$, $\overrightarrow{b}=2 \hat{i}+3 \hat{j}-\hat{k}$ and $\overrightarrow{c}=\lambda \hat{i}+\hat{j}+(2 \lambda-1) \hat{k}$. If $\overrightarrow{c}$ is parallel to the plane containing $\overrightarrow{a}$ and $\overrightarrow{b}$, then $\lambda$ is equal to

If $\overline{a}, \overline{b}$ and $\overline{c}$ are three non-coplanar vectors,then $(\overline{a}+\overline{b}+\overline{c}) \cdot[(\overline{a}+\overline{b}) \times(\overline{a}+\overline{c})]$ equals

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:

For what value of $\lambda$ are the vectors $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$,$\vec{b} = \lambda\hat{i} + 4\hat{j} + 7\hat{k}$,and $\vec{c} = -3\hat{i} - 2\hat{j} - 5\hat{k}$ coplanar?