The volume (in cubic units) of the tetrahedron with edges $\hat{i}+\hat{j}+\hat{k}$, $\hat{i}-\hat{j}+\hat{k}$ and $\hat{i}+2\hat{j}-\hat{k}$ is

The sum of the distinct real values of $\mu$,for which the vectors $\mu \hat{i} + \hat{j} + \hat{k}$,$\hat{i} + \mu \hat{j} + \hat{k}$,and $\hat{i} + \hat{j} + \mu \hat{k}$ are coplanar,is

If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$ and $\vec{c} = x\hat{i} + (x - 2)\hat{j} - \hat{k}$ are three vectors such that $\vec{c}$ lies in the plane of $\vec{a}$ and $\vec{b}$, then $x = ...$

Given the vectors $\vec x = 3i - 6j - k$,$\vec y = i + 4j - 3k$,and $\vec z = 3i - 4j - 12k$,find the projection of the vector $\vec x \times \vec y$ onto the vector $\vec z$.

If $2 \hat{i}-\hat{j}+3 \hat{k}$, $-12 \hat{i}-\hat{j}-3 \hat{k}$, $-\hat{i}+2 \hat{j}-4 \hat{k}$, and $\lambda \hat{i}+2 \hat{j}-\hat{k}$ are the position vectors of four coplanar points, then $\lambda=$

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: