In $\triangle ABC$,if $S$ is the circumcentre and $O$ is the orthocentre,then $\vec{OA} + \vec{OB} + \vec{OC} = $

Let $a$ and $b$ be linearly independent vectors such that $|a| = \sqrt{3}$, $|b| = 3$, and $|a - b| = 4$. If $a \times (2i + 2j - k) = (2i + 2j - k) \times b$ and $|(a + b) \cdot (3i + 4j + 2k)| = \sqrt{\lambda}$, then $\lambda = ...$

Let $\bar{a}, \bar{b}, \bar{c}$ be three vectors such that $\bar{a}+\bar{b}+\bar{c}=\bar{0}$,$|\bar{a}|=3$,$|\bar{b}|=4$,and $|\bar{c}|=5$. Then,find the value of $\bar{a} \cdot \bar{b}+\bar{b} \cdot \bar{c}+\bar{c} \cdot \bar{a}$.

$\hat{i} \cdot (\hat{k} \times \hat{j}) + \hat{j} \cdot (\hat{i} \times \hat{k}) + \hat{k} \cdot (\hat{i} \times \hat{j}) = \_\_\_\_$

If $\bar{a}=2 \hat{i}+3 \hat{j}+2 \hat{k}$,$\bar{b}=2 \hat{i}+\hat{j}-\hat{k}$ and $\bar{c}=3 \hat{i}-\hat{j}$ are such that $\bar{a}+\lambda \bar{b}$ is perpendicular to $\bar{c}$,then the value of $\lambda$ is

If the angle between the vectors $a$ and $b$ is $\theta$ and $a \cdot b = \cos \theta$,then the true statement is: