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

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:

If vectors $\vec{a}$ and $\vec{b}$ represent the sides $\vec{AB}$ and $\vec{BC}$ of a regular hexagon $ABCDEF$,find the vector represented by $\vec{FA}$.

If $\vec{a}+l \vec{b}+l^2 \vec{c}=0$ and $\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}=3(\vec{b} \times \vec{c})$,then the minimum value of such $l$ is

Observe the following lists. Then the correct match for List-$I$ from List-$II$ is:

List-$I$	List-$II$
$(A)$ $[\mathbf{a} \mathbf{b} \mathbf{c}]$	$1. |\mathbf{a}||\mathbf{b}|\cos(\mathbf{a}, \mathbf{b})$
$(B)$ $(\mathbf{c} \times \mathbf{a}) \times \mathbf{b}$	$2. (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$
$(C)$ $\mathbf{a} \times (\mathbf{b} \times \mathbf{c})$	$3. \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$
$(D)$ $\mathbf{a} \cdot \mathbf{b}$	$4. |\mathbf{a}||\mathbf{b}|$
	$5. (\mathbf{b} \cdot \mathbf{c})\mathbf{a} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$

Let $a, b, c$ be three non-coplanar vectors such that $r_1 = a - b + c$,$r_2 = b + c - a$,$r_3 = c + a + b$,and $r = 2a - 3b + 4c$. If $r = \lambda_1 r_1 + \lambda_2 r_2 + \lambda_3 r_3$,then: