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

Between the following two statements :
Statement $-I$ : Let $\vec{a}=\hat{i}+2\hat{j}-3\hat{k}$ and $\vec{b}=2\hat{i}+\hat{j}-\hat{k}$. Then the vector $\vec{r}$ satisfying $\vec{a} \times \vec{r}=\vec{a} \times \vec{b}$ and $\vec{a} \cdot \vec{r}=0$ is of magnitude $\sqrt{10}$.
Statement $-II$ : In a triangle $ABC$,$\cos 2A+\cos 2B+\cos 2C \geq -\frac{3}{2}$.

If $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}, \overrightarrow{b}=\hat{i}-\hat{j}+\hat{k}, \overrightarrow{c}=\hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{d}=\hat{i}-\hat{j}-\hat{k}$, then match the following List-$I$ with List-$II$:

List-$I$	List-$II$
$(i)$ $\overrightarrow{a} \cdot \overrightarrow{b}$	$(A)$ $\overrightarrow{a} \cdot \overrightarrow{d}$
(ii) $\overrightarrow{b} \cdot \overrightarrow{c}$	$(B)$ $3$
(iii) $[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]$	$(C)$ $\overrightarrow{b} \cdot \overrightarrow{d}$
(iv) $\overrightarrow{b} \times \overrightarrow{c}$	$(D)$ $2\hat{i}-2\hat{k}$
	$(E)$ $2\hat{j}+2\hat{k}$
	$(F)$ $4$

Let $\hat{i}, \hat{j}$ and $\hat{k}$ be the unit vectors along the three positive coordinate axes. Let $\vec{a}=3\hat{i}+\hat{j}-\hat{k}$,$\vec{b}=\hat{i}+b_2\hat{j}+b_3\hat{k}$ $(b_2, b_3 \in \mathbb{R})$,and $\vec{c}=c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$ $(c_1, c_2, c_3 \in \mathbb{R})$ be three vectors such that $b_2b_3 > 0$,$\vec{a} \cdot \vec{b} = 0$ and $\begin{bmatrix} 0 & -c_3 & c_2 \\ c_3 & 0 & -c_1 \\ -c_2 & c_1 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ b_2 \\ b_3 \end{bmatrix} = \begin{bmatrix} 3-c_1 \\ 1-c_2 \\ -1-c_3 \end{bmatrix}$. Then,which of the following is/are $TRUE$?

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:

The point $B$ divides the arc $AC$ of a quadrant of a circle in the ratio $1 : 2$. If $O$ is the centre and $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$,then the vector $\overrightarrow{OC}$ is