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 $\vec{a}$.

Let $a$ and $b$ be positive real numbers. Suppose $\overrightarrow{PQ} = a \hat{i} + b \hat{j}$ and $\overrightarrow{PS} = a \hat{i} - b \hat{j}$ are adjacent sides of a parallelogram $PQRS$. Let $\overrightarrow{u}$ and $\overrightarrow{v}$ be the projection vectors of $\overrightarrow{w} = \hat{i} + \hat{j}$ along $\overrightarrow{PQ}$ and $\overrightarrow{PS}$,respectively. If $|\vec{u}| + |\vec{v}| = |\vec{w}|$ and if the area of the parallelogram $PQRS$ is $8$,then which of the following statements is/are $TRUE$?
$(A)$ $a + b = 4$
$(B)$ $a - b = 2$
$(C)$ The length of the diagonal $PR$ of the parallelogram $PQRS$ is $4$
$(D)$ $\overrightarrow{w}$ is an angle bisector of the vectors $\overrightarrow{PQ}$ and $\overrightarrow{PS}$

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	$(4)$ $3$
	$(5)$ $5$

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:

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:

Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors in the $xyz$-space such that $\vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a} \neq 0$. If $A, B, C$ are points with position vectors $\vec{a}, \vec{b}, \vec{c}$ respectively,then the number of possible positions of the centroid of $\triangle ABC$ is