The position vectors of the vertices of a quadrilateral $ABCD$ are $a, b, c$ and $d$ respectively. The area of the quadrilateral formed by joining the midpoints of its sides is

Let $\vec{w}=\hat{i}+\hat{j}-2 \hat{k}$,and $\vec{u}$ and $\vec{v}$ be two vectors such that $\vec{u} \times \vec{v}=\vec{w}$ and $\vec{v} \times \vec{w}=\vec{u}$. Let $\alpha, \beta, \gamma$ and $t$ be real numbers such that $\vec{u}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$,$-t \alpha+\beta+\gamma=0$,$\alpha-t \beta+\gamma=0$,and $\alpha+\beta-t \gamma=0$. Match each entry in List-$I$ to the correct entry in List-$II$ and choose the correct option.

List-$I$	List-$II$
$(P)$ $|\vec{v}|^2$ is equal to	$(1)$ $0$
$(Q)$ If $\alpha=\sqrt{3}$,then $\gamma^2$ is equal to	$(2)$ $1$
$(R)$ If $\alpha=\sqrt{3}$,then $(\beta+\gamma)^2$ is equal to	$(3)$ $2$
$(S)$ If $\alpha=\sqrt{2}$,then $t+3$ is equal to	$(4)$ $3$
	$(5)$ $5$

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 $\vec{a} = 2\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$ be two vectors. Consider a vector $\vec{c} = \alpha\vec{a} + \beta\vec{b}$,where $\alpha, \beta \in \mathbb{R}$. If the projection of $\vec{c}$ on the vector $(\vec{a} + \vec{b})$ is $3\sqrt{2}$,then the minimum value of $(\vec{c} - (\vec{a} \times \vec{b})) \cdot \vec{c}$ is equal to:

Let $\overrightarrow{OA}=2 \hat{i}-3 \hat{j}+\hat{k}$,$\overrightarrow{OB}=\hat{i}-4 \hat{j}-3 \hat{k}$,and $\overrightarrow{OC}=-3 \hat{i}+\hat{j}+2 \hat{k}$ be the position vectors of three points $A$,$B$,and $C$ respectively. If $G$ is the centroid of triangle $ABC$,then $BC^2+CA^2+AB^2+9(OG)^2=$

If $\vec{a}, \vec{b}$ and $\vec{c}$ are non-coplanar vectors and if $\vec{d}$ is such that $\vec{d} = \frac{1}{x}(\vec{a} + \vec{b} + \vec{c})$ and $\vec{d} = \frac{1}{y}(\vec{b} + \vec{c} + \vec{d})$ where $x$ and $y$ are non-zero real numbers, then $\frac{1}{xy}(\vec{a} + \vec{b} + \vec{c} + \vec{d})$ is equal to: