The value of $[\vec{a}-\vec{b} \quad \vec{b}-\vec{c} \quad \vec{c}-\vec{a}]$ is equal to

Statement-$1$: Vectors $\vec{a}, \vec{b},$ and $\vec{c}$ are coplanar if and only if $\vec{a} \cdot (\vec{b} \times \vec{c}) = 0$.
Statement-$2$: Vectors $\vec{u}$ and $\vec{v}$ are perpendicular if and only if $\vec{u} \cdot \vec{v} = 0$,where $\vec{u} \times \vec{v}$ is a vector perpendicular to the plane of $\vec{u}$ and $\vec{v}$.

If $a = i + j + k$,$b = 4i + 3j + 4k$,and $c = i + \alpha j + \beta k$ are coplanar vectors and $|c| = \sqrt{3}$,then:

Let $a, b$ and $c$ be three vectors. Then the scalar triple product $[a, b, c]$ is equal to:

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

If $\vec{a}$ and $\vec{b}$ are mutually perpendicular unit vectors and $\vec{r}$ is a vector such that $\vec{r} \cdot \vec{a} = 0$,$\vec{r} \cdot \vec{b} = 1$,and $[\vec{r} \, \vec{a} \, \vec{b}] = 1$,then $\vec{r} = \dots$