A tetrahedron has vertices $\mathrm{O}(0,0,0), \mathrm{A}(1,2,1)$, $B(2,1,3), C(-1,1,2)$. Then the angle between the faces OAB and ABC will be

If $A=\left[\begin{array}{lll}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array}\right]_{3 \times 3}$, then $A^{-1}=$

The area bounded by the parabolas $y=9 x^2, y=\frac{x^2}{16}$ and the line $y=1$ is

$\int \frac{\mathrm{d} x}{3 \cos 2 x+5}$ equals