If $ABCDEF$ is a regular hexagon,and $\vec{AB} + \vec{AC} + \vec{AD} + \vec{AE} + \vec{AF} = k \vec{AD}$,then $k = \dots$

The vector that is parallel to the vector $2 \hat{i} - 2 \hat{j} - 4 \hat{k}$ and coplanar with the vectors $\hat{i} + \hat{j}$ and $\hat{j} + \hat{k}$ is

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 $ABC$,$\cos 2A+\cos 2B+\cos 2C \geq -\frac{3}{2}$.

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

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) $\overrightarrow{b} \times \overrightarrow{c}$	$(D)$ $2\hat{i}-2\hat{k}$
	$(E)$ $2\hat{j}+2\hat{k}$
	$(F)$ $4$

If the vector $\vec{a} = (x, y, z)$ makes an obtuse angle with the $y$-axis and makes equal angles with the vectors $\vec{b} = (y, -2z, 3x)$ and $\vec{c} = (2z, 3x, -y)$,and if $|\vec{a}| = 2\sqrt{3}$ and $\vec{a}$ is perpendicular to $\vec{d} = (1, -1, 2)$,find the vector $\vec{a}$.