The equation of the line passing through the point $(3,1,2)$ and perpendicular to the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\frac{x}{-3}=\frac{y}{2}=\frac{z}{5}$ is

Let $A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]$ and $B=\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]$ such that $\mathrm{AX}=\mathrm{B}$, then $\mathrm{X}=$

If $\bar{x}=\frac{\bar{b} \times \bar{c}}{[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]}, \bar{y}=\frac{\overline{\mathrm{c}} \times \overline{\mathrm{a}}}{[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]}$ and $\overline{\mathrm{z}}=\frac{\overline{\mathrm{a}} \times \overline{\mathrm{b}}}{[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]}$ where $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ are non-coplanar vectors, then value of $\bar{x} \cdot(\overline{\mathrm{a}}+\overline{\mathrm{b}})+\bar{y} \cdot(\overline{\mathrm{~b}}+\overline{\mathrm{c}})+\overline{\mathrm{z}} \cdot(\overline{\mathrm{c}}+\overline{\mathrm{a}})$ is

If in a triangle $A B C$, with usual notations, the angles are in A.P. and $b: c=\sqrt{3}: \sqrt{2}$, then angle $\mathrm{A}=$