In a triangle $A B C$, with usual notations, $3 \mathrm{~b}=\mathrm{a}+\mathrm{c}$, then $\cot \frac{\mathrm{A}}{2} \cdot \cot \frac{\mathrm{C}}{2}=$

$$ \int_0^{\frac{\pi}{2}} \frac{d x}{1+(\cot x)^{101}}= $$

The solution of $\left(1+y^2\right)+\left(x-\mathrm{e}^{\tan ^{-1} y}\right) \frac{\mathrm{d} y}{\mathrm{~d} x}=0$ is

If $\left[\begin{array}{lll}1 & 3 & 3 \\ 1 & 4 & 4 \\ 1 & 3 & 4\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}12 \\ 15 \\ 13\end{array}\right]$, then the value of $x^2+y^2+z^2=$