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:

$\text{If } \vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = 2\hat{i} - \hat{j} + 3\hat{k} \text{ and } \vec{c} = \hat{i} - \hat{j} \text{ and if } 6\hat{i} + 2\hat{j} + 3\hat{k} = \lambda_1(\vec{a} \times \vec{b}) + \lambda_2(\vec{b} \times \vec{c}) + \lambda_3(\vec{c} \times \vec{a}), \text{ then } (\lambda_1, \lambda_2, \lambda_3) = $

Let $\vec{a}$ and $\vec{b}$ be the vectors along the diagonals of a parallelogram having area $2 \sqrt{2}$. Let the angle between $\vec{a}$ and $\vec{b}$ be acute. Given $|\vec{a}|=1$ and $|\vec{a} \cdot \vec{b}|=|\vec{a} \times \vec{b}|$. If $\vec{c}=2 \sqrt{2}(\vec{a} \times \vec{b})-2 \vec{b}$,then find the angle between $\vec{b}$ and $\vec{c}$.

Let $\bar{a}$ and $\bar{b}$ be two vectors such that $|\bar{a}|=1$,$|\bar{b}|=4$,and $\bar{a} \cdot \bar{b}=2$. If $\bar{c}=(2 \bar{a} \times \bar{b})-3 \bar{b}$,then the angle between $\bar{b}$ and $\bar{c}$ is

The line passing through the point $i + 3j + 2k$ and perpendicular to the lines $r = (i + 2j - k) + \lambda (2i + j + k)$ and $r = (2i + 6j + k) + \mu (i + 2j + 3k)$ is:

Let $\vec{b}=\hat{i}+\hat{j}+\lambda \hat{k}, \lambda \in R$. If $\vec{a}$ is a vector such that $\vec{a} \times \vec{b}=13 \hat{i}-\hat{j}-4 \hat{k}$ and $\vec{a} \cdot \vec{b}+21=0$,then $(\vec{b}-\vec{a}) \cdot(\hat{k}-\hat{j})+(\vec{b}+\vec{a}) \cdot(\hat{i}-\hat{k})$ is equal to