Let $\vec{a} = 2\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$ be two vectors. Consider a vector $\vec{c} = \alpha\vec{a} + \beta\vec{b}$,where $\alpha, \beta \in \mathbb{R}$. If the projection of $\vec{c}$ on the vector $(\vec{a} + \vec{b})$ is $3\sqrt{2}$,then the minimum value of $(\vec{c} - (\vec{a} \times \vec{b})) \cdot \vec{c}$ 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 $\hat{a}, \hat{b}$ and $\hat{c}$ are non-coplanar vectors and if $\hat{d}$ is such that $\hat{d} = \frac{1}{x}(\hat{a} + \hat{b} + \hat{c})$ and $\hat{d} = \frac{1}{y}(\hat{b} + \hat{c} + \hat{d})$ where $x$ and $y$ are non-zero real numbers,then $\frac{1}{xy}(\hat{a} + \hat{b} + \hat{c} + \hat{d})$ equals 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 position vectors of coplanar points $A, B, C,$ and $D$ are $\vec{a}, \vec{b}, \vec{c},$ and $\vec{d}$ respectively,such that $(\vec{a} - \vec{d}) \cdot (\vec{b} - \vec{c}) = 0$ and $(\vec{b} - \vec{d}) \cdot (\vec{c} - \vec{a}) = 0.$ Then the point $D$ of the triangle $ABC$ is

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

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