Let $O$ be the origin, $\overrightarrow{O P}=\vec{a}$ and $\overrightarrow{O Q}=\vec{b}$. If $R$ is the point on $\overrightarrow{O P}$ such that $\overrightarrow{O P}=5 \overrightarrow{O R}$, and $M$ is the point such that $\overrightarrow{O Q}=5 \overrightarrow{R M}$, then $\overrightarrow{P M}$ is equal to :

Let $\vec{a}=\sqrt{7} \hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{j}+2 \hat{k}$. If $\vec{r}$ is a vector such that $\vec{r} \times \vec{a}+\vec{a} \times \vec{b}=\overrightarrow{0}$ and $\vec{r} \cdot \vec{a}=0$, then $|3 \vec{r}|^2$ is equal to:

Let $\hat{u}$ and $\hat{v}$ be unit vectors inclined at an acute angle such that $|\hat{u} \times \hat{v}|=\frac{\sqrt{3}}{2}$. If $\overrightarrow{\mathrm{A}}=\lambda \hat{u}+\hat{v}+(\hat{u} \times \hat{v})$, then $\lambda$ is equal to:

Let the vectors $\vec{a} = -\hat{i} + \hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} + 3\hat{j} + \hat{k}$. For some $\lambda, \mu \in \mathbb{R}$, let $\vec{c} = \lambda \vec{a} + \mu \vec{b}$.

If $\vec{c} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) = 10$ and $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = -2$, then $|\vec{c}|^2$ is equal to :