The position vectors of coplanar points $A, B, C,$ and $D$ are $\vec{a}, \vec{b}, \vec{c},$ and $\vec{d}$ respectively,such that $(\vec{a} - \vec{d}) \cdot (\vec{b} - \vec{c}) = 0$ and $(\vec{b} - \vec{d}) \cdot (\vec{c} - \vec{a}) = 0.$ Then the point $D$ of the triangle $ABC$ is

Let $\vec{a}$ and $\vec{b}$ be two unit vectors such that $|\vec{a} + \vec{b}| = \sqrt{3}$. If $\vec{c} = \vec{a} + 2\vec{b} + 3(\vec{a} \times \vec{b})$,then $2|\vec{c}|$ is equal to

In a quadrilateral $ABCD$, the point $P$ divides $DC$ in the ratio $1:2$ and $Q$ is the midpoint of $AC$. If $\overrightarrow{AB}+2\overrightarrow{AD}+\overrightarrow{BC}-2\overrightarrow{DC}=k\overrightarrow{PQ}$, then $k$ is equal to

The points $D, E, F$ divide $BC, CA$ and $AB$ of the triangle $ABC$ in the ratio $1:4, 3:2$ and $3:7$ respectively and the point $K$ divides $AB$ in the ratio $1:3$,then $(\overrightarrow{AD} + \overrightarrow{BE} + \overrightarrow{CF}) : \overrightarrow{CK}$ is equal to

The vector $\frac{1}{3}(2i - 2j + k)$ is

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