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

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

The minimum value of $(x_1 - x_2)^2 + (\sqrt{2 - x_1^2} - \frac{9}{x_2})^2$ where $x_1 \in (0, \sqrt{2})$ and $x_2 \in R^+$.

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

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

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three unit vectors such that $\vec{a}+\vec{b}+\vec{c}=\vec{0}$. If $\lambda=\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$ and $\vec{d}=\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}$,then the ordered pair $(\lambda, \vec{d})$ is equal to:

Given that $a$ and $b$ are two unit non-collinear vectors,if $u = a - (a \cdot b)b$ and $v = a \times b$,then find $|v| =$.