In a triangle $A B C$, with usual notations, $\cot \left(\frac{A+B}{2}\right) \cdot \tan \left(\frac{A-B}{2}\right)=$

In a triangle ABC , with usual notations, $(\mathrm{a}+\mathrm{b}+\mathrm{c})(\mathrm{a}+\mathrm{b}-\mathrm{c})=3 \mathrm{ab}$, then $\angle \mathrm{C}=$

If $\bar{a}, \bar{b}, \bar{c}$ are three coplanar vectors such that $|\overline{\mathrm{a}}|=1,|\overline{\mathrm{~b}}|=2, \overline{\mathrm{~b}} \cdot \overline{\mathrm{c}}=8$, the angle between $\overline{\mathrm{b}}$ and $\overline{\mathrm{c}}$ is $45^{\circ}$, then $|\overline{\mathrm{a}} \times(\overline{\mathrm{b}} \times \overline{\mathrm{c}})|=$

Let $\overline{\mathrm{a}}, \overline{\mathrm{b}}$ and $\overline{\mathrm{c}}$ be vectors of magnitude 2,3 and 4 respectively. If $\bar{a}$ is perpendicular to $(\overline{\mathrm{b}}+\overline{\mathrm{c}}), \overline{\mathrm{b}}$ is perpendicular to ( $\overline{\mathrm{c}}+\overline{\mathrm{a}}$ ) and $\overline{\mathrm{c}}$ is perpendicular to $(\bar{a}+\bar{b})$, then the magnitude of $\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}$ is