If $\vec{a} = \lambda x \hat{i} + y \hat{j} + 4z \hat{k}$,$\vec{b} = y \hat{i} + x \hat{j} + 3y \hat{k}$,and $\vec{c} = -z \hat{i} - 2z \hat{j} - (\lambda + 1) \hat{k}$ are the sides of the triangle $ABC$,where $x, y, z$ are not all zero,such that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$,then the value of $\lambda$ is:

Let $\vec{a}_n = (\tan \theta_n)\hat{i} + \hat{j}$ and $\vec{b}_n = \hat{i} - (\cot \theta_n)\hat{j}$, where $\theta_n = \frac{2^{n-1}\pi}{2^n+1}$, for some $n \in N, n > 5$. Then the value of $\frac{\sum_{k=1}^n |\vec{a}_k|^2}{\sum_{k=1}^n |\vec{b}_k|^2}$ is . . . . . . .

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

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-zero vectors such that $\vec{b}$ and $\vec{c}$ are non-collinear. If $\vec{a}+5\vec{b}$ is collinear with $\vec{c}$,$\vec{b}+6\vec{c}$ is collinear with $\vec{a}$,and $\vec{a}+\alpha\vec{b}+\beta\vec{c}=\vec{0}$,then $\alpha+\beta$ is equal to

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

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

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