If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = \hat{i} - \hat{j}$ and $\vec{c}$ are three vectors such that $\vec{a} \times \vec{c} = \vec{b}$ and $\vec{a} \cdot \vec{c} = 3$, then $\vec{c} \cdot (\vec{a} - 2\vec{b})$ is equal to . . . . . . .

If $\vec{a}, \vec{b}, \vec{c}$ are three non-coplanar vectors and $\vec{r}$ is any vector,then $[\vec{b} \, \vec{c} \, \vec{r}] \vec{a} + [\vec{c} \, \vec{a} \, \vec{r}] \vec{b} + [\vec{a} \, \vec{b} \, \vec{r}] \vec{c} = \dots$

$ [\vec{a}+2 \vec{b}-\vec{c}, \vec{a}-\vec{b}, \vec{a}-\vec{b}-\vec{c}] $

If the four points $A(6,2,4)$,$B(1,3,5)$,$C(1,-2,3)$,and $D(6, k, 2)$ are coplanar,then $k=$

Let $a = \hat{i} - 2\hat{j} + 3\hat{k}$ and $b = 2\hat{i} + \hat{j} + \hat{k}$. If $c$ is a unit vector such that $[a \ b \ c]$ is maximum, then $c =$

$\vec{a}$ is a vector perpendicular to the plane containing non-zero vectors $\vec{b}$ and $\vec{c}$. If $\vec{a}, \vec{b}, \vec{c}$ are such that $|\vec{a}+\vec{b}+\vec{c}|=\sqrt{|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2}$, then $|(\vec{a} \times \vec{b}) \cdot \vec{c}|+|(\vec{a} \times \vec{b}) \times \vec{c}|=$