$$\frac{\mathrm{d}}{\mathrm{~d} x}\left(\cos ^{-1}\left(\frac{x-\frac{1}{x}}{x+\frac{1}{x}}\right)\right)=$$

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

The approximate value of $\tan ^{-1}(0.999)$ is (use $\pi=3.1415$ )

Let P be a plane passing through the points $(2,1,0),(4,1,1)$ and $(5,0,1)$ and $R$ be the point $(2,1,6)$. Then image of $R$ in the plane $P$ is