If $a \perp b$ and $(a+b) \perp (a+mb)$,then $m$ is equal to

Let the position vectors of the vertices of a triangle $ABC$ be $\bar{a}, \bar{b}, \bar{c}$. If on the plane of the triangle,$P$ is a point having position vector $\bar{x}$ such that $\bar{x} \cdot (\bar{c} - \bar{b}) = \bar{a} \cdot \bar{c} - \bar{a} \cdot \bar{b}$ and $\bar{x} \cdot (\bar{a} - \bar{c}) = \bar{a} \cdot \bar{b} - \bar{b} \cdot \bar{c}$,then for the triangle $ABC$,$P$ is the

If $\overline{a}, \overline{b}, \overline{c}$ are non-coplanar vectors and $\overline{p}=\frac{\overline{b} \times \overline{c}}{[\overline{a} \overline{b} \overline{c}]}, \overline{q}=\frac{\overline{c} \times \overline{a}}{[\overline{a} \overline{b} \overline{c}]}, \overline{r}=\frac{\overline{a} \times \overline{b}}{[\overline{a} \overline{b} \overline{c}]}, \quad$ then $2 \overline{a} \cdot \overline{p}+\overline{b} \cdot \overline{q}+\overline{c} \cdot \overline{r}=$

If $|a| = 3$,$|b| = 4$ and the angle between $a$ and $b$ is $120^\circ$,then $|4a + 3b| = $

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

If $A=(0,4,-3)$,$B=(5,0,12)$,and $C=(7,24,0)$,then $\angle BAC=$