Let $\overrightarrow{\mathrm{a}}=4 \hat{i}-\hat{j}+3 \hat{k}, \overrightarrow{\mathrm{~b}}=10 \hat{i}+2 \hat{j}-\hat{k}$ and a vector $\overrightarrow{\mathrm{c}}$ be such that $2(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})+3(\overrightarrow{\mathrm{~b}} \times \overrightarrow{\mathrm{c}})=\overrightarrow{0}$.

If $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=15$, then $\overrightarrow{\mathrm{c}} \cdot(\hat{i}+\hat{j}-3 \hat{k})$ is equal to :

Let $\overrightarrow{\mathrm{a}}=2 \hat{i}+3 \hat{j}+3 \hat{k}$ and $\overrightarrow{\mathrm{b}}=6 \hat{i}+3 \hat{j}+3 \hat{k}$. Then the square of the area of the triangle with adjacent sides determined by the vectors $(2 \vec{a}+3 \vec{b})$ and $(\vec{a}-\vec{b})$ is :

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: