If $\overrightarrow{a}+2 \overrightarrow{b}+3 \overrightarrow{c}=\overrightarrow{0}$,then $\overrightarrow{a} \times \overrightarrow{b}+\overrightarrow{b} \times \overrightarrow{c}+\overrightarrow{c} \times \overrightarrow{a}$ is equal to

Area of a rectangle having vertices $A, B, C$ and $D$ with position vectors $-\hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}, \hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}, \hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}$ and $-\hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}$ respectively is

$\vec{u}, \vec{v}, \vec{w}$ are three unit vectors. Let $\vec{p}=\vec{u}+\vec{v}+\vec{w}$ and $\vec{q}=\vec{u} \times(\vec{v} \times \vec{w})$. If $\vec{p} \cdot \vec{u}=\frac{3}{2}, \vec{p} \cdot \vec{v}=\frac{7}{4}, |\vec{p}|=2$ and $\vec{v}=K \vec{q}$,then $K=$

If $\bar{a}=2 \hat{i}+3 \hat{j}-\hat{k}, \bar{b}=-\hat{i}+2 \hat{j}-4 \hat{k}$ and $\bar{c}=\hat{i}+\hat{j}-2 \hat{k}$,then $(\bar{a} \times \bar{b}) \cdot(\bar{a} \times \bar{c})=$

Let $\overrightarrow{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}$.
Assertion $(A)$ : The identity $|\overrightarrow{a} \times \hat{i}|^2+|\overrightarrow{a} \times \hat{j}|^2+|\overrightarrow{a} \times \hat{k}|^2=2|\overrightarrow{a}|^2$ holds for $\overrightarrow{a}$.
Reason $(R)$ : $\overrightarrow{a} \times \hat{i}=a_3 \hat{j}-a_2 \hat{k}$,$\overrightarrow{a} \times \hat{j}=a_1 \hat{k}-a_3 \hat{i}$,and $\overrightarrow{a} \times \hat{k}=a_2 \hat{i}-a_1 \hat{j}$.
Which of the following is correct?

Let $O$ be the origin and the position vector of the point $P$ be $-\hat{i}-2\hat{j}+3\hat{k}$. If the position vectors of the points $A, B$ and $C$ are $-2\hat{i}+\hat{j}-3\hat{k}$,$2\hat{i}+4\hat{j}-2\hat{k}$ and $-4\hat{i}+2\hat{j}-\hat{k}$ respectively,then the projection of the vector $\overline{OP}$ on a vector perpendicular to the vectors $\overline{AB}$ and $\overline{AC}$ is $......$.