If the plane $\frac{x}{3}+\frac{y}{2}-\frac{z}{4}=1$ cuts the co-ordinate axes at points $\mathrm{A}, \mathrm{B}$ and C , then the area of the triangle ABC is

If $\tan ^{-1}\left(\frac{x}{2}\right)+\tan ^{-1}\left(\frac{y}{2}\right)+\tan ^{-1}\left(\frac{z}{2}\right)=\frac{\pi}{2} \quad$ then $x y+y z+z x=$

In a triangle ABC , with usual notations if $\frac{2 \cos \mathrm{~A}}{\mathrm{a}}+\frac{\cos \mathrm{B}}{\mathrm{b}}+\frac{2 \cos \mathrm{C}}{\mathrm{c}}=\frac{\mathrm{a}}{\mathrm{bc}}+\frac{\mathrm{b}}{\mathrm{ca}}$ then $\angle \mathrm{A}=$

If $f(1)=1, f^{\prime}(1)=3$, then the derivative of $\mathrm{f}(\mathrm{f}(\mathrm{f}(x)))+(\mathrm{f}(x))^2$ at $x=1$ is