For a triangle $ABC$, let $\vec{p}=\vec{BC}$, $\vec{q}=\vec{CA}$ and $\vec{r}=\vec{BA}$. If $|\vec{p}|=2\sqrt{3}$, $|\vec{q}|=2$ and $\cos \theta = \frac{1}{\sqrt{3}}$ where $\theta$ is the angle between $\vec{p}$ and $\vec{q}$, then $|\vec{p} \times (\vec{q}-3\vec{r})|^{2}+3|\vec{r}|^{2}$ 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

Let $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ be the position vectors of the vertices $A, B, C$ respectively of $\triangle ABC$. The vector area of $\triangle ABC$ is:

If $\overrightarrow{a}=2 \hat{i}-5 \hat{j}+8 \hat{k}$ and $\overrightarrow{b}=7 \hat{i}-5 \hat{j}+3 \hat{k}$ are two vectors and $(2 \overrightarrow{a}-3 \overrightarrow{b}) \times(4 \overrightarrow{a}+\overrightarrow{b})=x \hat{i}+y \hat{j}+z \hat{k}$,then $x+y+z=$

Let $\bar{a} = \hat{i} + 2\hat{j} - 2\hat{k}$ and $\bar{b} = \hat{i} - \hat{j} + \hat{k}$. If $\bar{c}$ is a vector such that $\bar{a} \cdot \bar{c} = |\bar{c}|$,$|\bar{c} - \bar{a}| = 2\sqrt{2}$ and the angle between $\bar{a} \times \bar{b}$ and $\bar{c}$ is $60^{\circ}$,then $|(\bar{a} \times \bar{b}) \times \bar{c}|$ is equal to

$A$ non-zero vector $\vec{a}$ is parallel to the line of intersection of the plane determined by the vectors $\hat{i}$ and $\hat{i}+\hat{j}$ and the plane determined by vectors $\hat{i}-\hat{j}$ and $\hat{i}+\hat{k}$. The angle between $\vec{a}$ and $(\hat{i}-2\hat{j}+2\hat{k})$ is