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) = $

Using vectors,find the area of the triangle $ABC$ with vertices $A(1, 2, 3)$,$B(2, -1, 4)$ and $C(4, 5, -1)$.

If $u = 2i + 2j - k$ and $v = 6i - 3j + 2k,$ then a unit vector perpendicular to both $u$ and $v$ is

The vector of magnitude $6$ units and perpendicular to vectors $2 \hat{i}+\hat{j}-3 \hat{k}$ and $\hat{i}-2 \hat{j}+\hat{k}$ is

Let $\overrightarrow{a}$ be a non-zero vector parallel to the line of intersection of the two planes passing through the origin and containing the vectors $(\hat{i}+\hat{j}, \hat{i}+\hat{k})$ and $(\hat{i}-\hat{j}, \hat{j}-\hat{k})$ respectively. If $\theta$ is the angle between the vector $\vec{a}$ and the vector $\vec{b}=2\hat{i}-2\hat{j}+\hat{k}$ and $\vec{a} \cdot \vec{b}=6$,then the ordered pair $(\theta, |\vec{a} \times \vec{b}|)$ is equal to