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 the four points with position vectors $-2\hat{i}+\hat{j}+\hat{k}$, $\hat{i}+\hat{j}+\hat{k}$, $\hat{j}-\hat{k}$, and $\lambda\hat{j}+\hat{k}$ are coplanar, then $\lambda$ is equal to

If $\overrightarrow x = 3\hat i - 6\hat j - \hat k$,$\overrightarrow y = \hat i + 4\hat j - 3\hat k$ and $\overrightarrow z = 3\hat i - 4\hat j - 12\hat k$,then the magnitude of the projection of $\overrightarrow x \times \overrightarrow y$ on $\overrightarrow z$ is

$a \cdot [(b + c) \times (a + b + c)]$ is equal to

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 $[(\overline{a}+2 \overline{b}+3 \overline{c}) \times(\overline{b}+2 \overline{c}+3 \overline{a})] \cdot(\overline{c}+2 \overline{a}+3 \overline{b})=54$,then the value of $[\overline{a} \ \overline{b} \ \overline{c}]$ is