If $ABCD$ is a quadrilateral,then the resultant force represented by $\vec{BA}, \vec{BC}, \vec{CD}$ and $\vec{DA}$ is equal to:

Let $\vec{a}=\hat{i}+\hat{j}+2\hat{k}$,$\vec{b}=2\hat{i}-3\hat{j}+\hat{k}$,and $\vec{c}=\hat{i}-\hat{j}+\hat{k}$ be three given vectors. Let $\vec{v}$ be a vector in the plane of $\vec{a}$ and $\vec{b}$ whose projection on $\vec{c}$ is $\frac{2}{\sqrt{3}}$. If $\vec{v} \cdot \hat{j}=7$,then $\vec{v} \cdot (\hat{i}+\hat{k})$ is equal to

If $\bar{a}, \bar{b}, \bar{c}$ are three vectors such that $|\bar{a}|=\sqrt{31}, 4|\bar{b}|=|\bar{c}|=2$ and $2(\bar{a} \times \bar{b})=3(\bar{c} \times \bar{a})$ and if the angle between $\bar{b}$ and $\bar{c}$ is $\frac{2\pi}{3}$,then $\left|\frac{\bar{a} \times \bar{c}}{\bar{a} \cdot \bar{b}}\right|^2=$

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

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

If $a=2 \hat{i}+3 \hat{j}+\hat{k}$, $b=\hat{i}-3 \hat{j}-5 \hat{k}$ and $c=3 \hat{i}-4 \hat{k}$, then match the items of List-$I$ with those of List-$II$.

$A$. Unit vector in the direction opposite to that $a-b$ is	$(i) \ 5 \hat{i} + 3 \hat{j} - 3 \hat{k}$
$B$. If $\vec{AB} = a, \vec{BC} = b$, then $\vec{CA} =$	$(ii) \ 2 \hat{i} - \frac{8}{3} \hat{k}$
$C$. If $a, b, c$ are the position vectors of the vertices of a triangle then, its centroid is	$(iii) \ -3 \hat{i} + 4 \hat{k}$
$D$. If $d$ is a vector of magnitude $2 \sqrt{14}$ and parallel to the vector $a$, then $b + d =$	$(iv) \ -\frac{\hat{i}}{\sqrt{73}} - \frac{6 \hat{j}}{\sqrt{73}} - \frac{6 \hat{k}}{\sqrt{73}}$
	$(v) \ 3 \hat{i} + 5 \hat{j} - 3 \hat{k}$

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: