In the given figure,if a vector $x$ satisfies the equation $x - w = v$,then $x = ?$

Let $\vec{a}$ and $\vec{b}$ be two unit vectors such that $|\vec{a} + \vec{b}| = \sqrt{3}$. If $\vec{c} = \vec{a} + 2\vec{b} + 3(\vec{a} \times \vec{b})$,then $2|\vec{c}|$ is equal to

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

The points $D, E, F$ divide $BC, CA$ and $AB$ of the triangle $ABC$ in the ratio $1:4, 3:2$ and $3:7$ respectively and the point $K$ divides $AB$ in the ratio $1:3$,then $(\overrightarrow{AD} + \overrightarrow{BE} + \overrightarrow{CF}) : \overrightarrow{CK}$ is equal to

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 $\vec{a}$.

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: