If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$,$\vec{b}=\hat{i}-\hat{j}+2\hat{k}$,$\vec{c}=x\hat{i}+(x-2)\hat{j}-\hat{k}$ and $\vec{c}$ is a linear combination of $\vec{a}$ and $\vec{b}$,then the value of $x$ is:

If $\bar{a}, \bar{b}$ and $\bar{c}$ are any three non-zero vectors,then $(\bar{a}+2 \bar{b}+\bar{c}) \cdot[(\bar{a}-\bar{b}) \times(\bar{a}-\bar{b}-\bar{c})]=$

Let $\vec{u} = a\hat{i} + b\hat{j} + c\hat{k}$,$\vec{v} = b\hat{i} + c\hat{j} + a\hat{k}$,and $\vec{w} = c\hat{i} + a\hat{j} + b\hat{k}$. If $[\vec{u} \, \vec{v} \, \vec{w}] = 0$ and $\vec{w} = \lambda \vec{x} + \mu \vec{y}$ where $(a + b + c) \neq 0$ and $\lambda, \mu \neq 0$,then the vectors $\vec{x}, \vec{y}, \vec{u}, \vec{v}, \vec{w}$ are:

If $a = -3i + 7j + 5k$,$b = -3i + 7j - 3k$,and $c = 7i - 5j - 3k$ are the three coterminous edges of a parallelepiped,then its volume is

For any non-zero vectors $a, b, c$,$a \cdot[(b+c) \times(a+b+c)] = \ldots .$

Let the volume of a parallelepiped whose coterminous edges are given by $\overrightarrow{u}=\hat{i}+\hat{j}+\lambda \hat{k}$,$\overrightarrow{v}=\hat{i}+\hat{j}+3 \hat{k}$ and $\overrightarrow{w}=2 \hat{i}+\hat{j}+\hat{k}$ be $1 \text{ cu. unit}$. If $\theta$ is the angle between the edges $\overrightarrow{u}$ and $\overrightarrow{w}$,then $\cos \theta$ can be