Let the position vectors of three vertices of a triangle be $4 \vec{p}+\vec{q}-3 \vec{r},-5 \vec{p}+\vec{q}+2 \vec{r}$ and $2 \vec{p}-\vec{q}+2 \vec{r}$. If the position vectors of the orthocenter and the circumcenter of the triangle are $\frac{\vec{p}+\vec{q}+\vec{r}}{4}$ and $\alpha \vec{p}+\beta \vec{q}+\gamma \vec{r}$ respectively, then $\alpha+2 \beta+5 \gamma$ is equal to :

Let $\overrightarrow{\mathrm{a}}=3 \hat{i}-\hat{j}+2 \hat{k}, \overrightarrow{\mathrm{~b}}=\overrightarrow{\mathrm{a}} \times(\hat{i}-2 \hat{k})$ and $\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}} \times \hat{k}$. Then the projection of $\overrightarrow{\mathrm{c}}-2 \hat{j}$ on $\vec{a}$ is :

Let $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=3 \hat{i}+\hat{j}-\hat{k}$ and $\vec{c}$ be three vectors such that $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$. If the vector $\vec{C}$ is perpendicular to $\vec{b}$ and $\vec{a} \cdot \vec{c}=5$, then $|\vec{c}|$ is equal to

Let the point A divide the line segment joining the points $\mathrm{P}(-1,-1,2)$ and $\mathrm{Q}(5,5,10)$ internally in the ratio $r: 1(r>0)$. If O is the origin and $(\overrightarrow{\mathrm{OQ}} \cdot \overrightarrow{\mathrm{OA}})-\frac{1}{5}|\overrightarrow{\mathrm{OP}} \times \overrightarrow{\mathrm{OA}}|^2=10$, then the value of r is :