Let $u, v$ and $w$ be three vectors such that $u+v+w=0$,$|u|=3$,$|v|=5$ and $|w|=7$. Then the angle between $u$ and $v$ is (in $^{\circ}$)

Let $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$. If $\vec{b}$ is a vector such that $\vec{a} \cdot \vec{b} = |\vec{b}|^2$ and $|\vec{a} - \vec{b}| = \sqrt{7}$,then find $|\vec{b}|$.

Let $(x, y) \in (R \times R)$ and $\vec{a} = x \hat{i} + 2 \hat{j} - \hat{k}$, $\vec{b} = 6 \hat{i} - y \hat{j} + 2 \hat{k}$ be two vectors. If $|\vec{a} \times \vec{b}|^2 + |\vec{a} \cdot \vec{b}|^2 = f(x) g(y)$, then $f(x) + g(y) - 46 = 0$ represents:

Let $\vec{\alpha}=\hat{i}+\hat{j}+\hat{k}$, $\vec{\beta}=\hat{i}-\hat{j}-\hat{k}$ and $\vec{\gamma}=-\hat{i}+\hat{j}-\hat{k}$ be three vectors. $A$ vector $\vec{\delta}$, in the plane of $\vec{\alpha}$ and $\vec{\beta}$, whose projection on $\vec{\gamma}$ is $\frac{1}{\sqrt{3}}$, is given by

$3 \hat{i}-2 \hat{j}-\hat{k}, -2 \hat{i}-\hat{j}+3 \hat{k}$ and $-\hat{i}+3 \hat{j}-2 \hat{k}$ are the position vectors of the vertices $A, B$ and $C$ of a $\triangle ABC$ respectively. If $H$ is its orthocenter,then $\overrightarrow{HA}+\overrightarrow{HB}+\overrightarrow{HC} = $

If $|\bar{a} \times \bar{b}|^2+(\bar{a} \cdot \bar{b})^2=144$ and $|\bar{a}|=4$,then $|\bar{b}|=$