Let $\vec{a}=-\hat{i}+\hat{j}+2\hat{k}$, $\vec{b}=\hat{i}-\hat{j}-3\hat{k}$, $\vec{c}=\vec{a}\times\vec{b}$ and $\vec{d}=\vec{c}\times\vec{a}$. Then $(\vec{a}-\vec{b}) \cdot \vec{d}$ is equal to :

$\hat{a}, \hat{b}$,and $\hat{c}$ are three unit vectors such that $\hat{a} \times(\hat{b} \times \hat{c})=\frac{\sqrt{3}}{2}(\hat{b}+\hat{c})$. If $\hat{b}$ is not parallel to $\hat{c}$,then the angle between $\hat{a}$ and $\hat{b}$ is

If $(\vec{a} \times \vec{b}) \times \vec{c} = \vec{a} \times (\vec{b} \times \vec{c})$,where $\vec{a}, \vec{b},$ and $\vec{c}$ are any three vectors such that $\vec{a} \cdot \vec{b} \neq 0$ and $\vec{b} \cdot \vec{c} \neq 0$,then $\vec{a}$ and $\vec{c}$ are:

If $a = i + j - k$,$b = i - j + k$,and $c = i - j - k$,then $a \times (b \times c) = \dots$

If $\vec{x} \cdot \vec{y} = 0$ then,$(\vec{y} \times \vec{x}) \times \vec{x} = $ . . . . . . . where,$|\vec{x}| = 1$.

Let $\vec{a}=-\hat{i}-\hat{j}+\hat{k}$,$\vec{a} \cdot \vec{b}=1$ and $\vec{a} \times \vec{b}=\hat{i}-\hat{j}$. Then $\vec{a}-6 \vec{b}$ is equal to