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

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 $\vec{a}=\hat{i}+\hat{j}+\hat{k}$,$\vec{b}=2\hat{i}+2\hat{j}+\hat{k}$ and $\vec{d}=\vec{a} \times \vec{b}$. If $\vec{c}$ is a vector such that $\vec{a} \cdot \vec{c}=|\vec{c}|$,$|\vec{c}-2\vec{a}|^2=8$ and the angle between $\vec{d}$ and $\vec{c}$ is $\frac{\pi}{4}$,then $|10-3\vec{b} \cdot \vec{c}|+|\vec{d} \times \vec{c}|^2$ is equal to . . . . . .

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

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

Let $a, b, c$ be three non-coplanar vectors such that $r_1 = a - b + c$,$r_2 = b + c - a$,$r_3 = c + a + b$,and $r = 2a - 3b + 4c$. If $r = \lambda_1 r_1 + \lambda_2 r_2 + \lambda_3 r_3$,then:

The position vectors of the vertices of a quadrilateral $ABCD$ are $a, b, c$ and $d$ respectively. The area of the quadrilateral formed by joining the midpoints of its sides is