$\vec{a}=\hat{i}+\hat{j}-2 \hat{k}$, $\vec{b}=\hat{i}-2 \hat{j}+\hat{k}$ and $\vec{c}=2 \hat{i}+\hat{j}-\hat{k}$ are three vectors. If $\vec{d}$ is a normal to the plane of $\vec{a}$ and $\vec{b}$ and $\vec{d} \cdot \vec{c}=2$, then $|\vec{d}|=$

$|\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = $

If $a, b, c$ are position vectors of vertices of a triangle $ABC$,then the unit vector perpendicular to its plane is:

Let $\vec{p}=2 \hat{i}+3 \hat{j}+\hat{k}$ and $\vec{q}=\hat{i}+2 \hat{j}+\hat{k}$ be two vectors. If a vector $\vec{r}=(\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k})$ is perpendicular to each of the vectors $(\vec{p}+\vec{q})$ and $(\vec{p}-\vec{q})$,and $|\vec{r}|=\sqrt{3}$,then $|\alpha|+|\beta|+|\gamma|$ is equal to $.....$

If the vectors $a$ and $b$ are mutually perpendicular,then $a \times \{ a \times \{ a \times (a \times b)\} \}$ is equal to

If $A, B, C, D$ are any four points in space, then $|\overrightarrow{AB} \times \overrightarrow{CD} + \overrightarrow{BC} \times \overrightarrow{AD} + \overrightarrow{CA} \times \overrightarrow{BD}|$ is equal to (where $\Delta$ denotes the area of $\Delta ABC$) (in $\Delta$)