If the vectors $(1 - x)\hat i + \hat j + \hat k$,$\hat i + (1 - y)\hat j + \hat k$,and $\hat i + \hat j + (1 - z)\hat k$ are coplanar,then the value of $\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$ is $(x, y, z \neq 0)$.

$\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are non-coplanar vectors such that $\overrightarrow{P} = \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}$,$\overrightarrow{Q} = 4\overrightarrow{a} + 3\overrightarrow{b} + 4\overrightarrow{c}$,and $\overrightarrow{R} = \overrightarrow{a} + \alpha\overrightarrow{b} + \beta\overrightarrow{c}$ are linearly dependent vectors. Then,the number of possible values of $\alpha$ is:

Let $a, b, c$ be distinct non-negative numbers. If the vectors $a\hat{i} + a\hat{j} + c\hat{k}$,$\hat{i} + \hat{k}$,and $c\hat{i} + c\hat{j} + b\hat{k}$ lie in a plane,then $c$ is

$[\hat{i}-\hat{j}, \hat{j}-\hat{k}, \hat{k}-\hat{i}]$ is equal to

Let $\vec{a} = \hat{i} - \hat{k}$,$\vec{b} = x\hat{i} + \hat{j} + (1 - x)\hat{k}$,and $\vec{c} = y\hat{i} + x\hat{j} + (1 + x - y)\hat{k}$. Then the scalar triple product $[\vec{a} \, \vec{b} \, \vec{c}]$ depends on:

If $\overline{u}, \overline{v}$ and $\overline{w}$ are three non-coplanar vectors,then $(\bar{u}+\bar{v}-\bar{w}) \cdot [(\bar{u}-\bar{v}) \times (\bar{v}-\bar{w})]$ is equal to