I2.1 If $x, y$ and $z$ are positive real numbers such that $\frac{x+y-z}{z}=\frac{x-y+z}{y}=\frac{-x+y+z}{x}$ and $a=\frac{(x+y) \cdot(y+z) \cdot(z+x)}{x y z}$, find the value of $a$.

I2.2 Let $u$ and $t$ be positive integers such that $u+t+u t=4 a+2$. If $b=u+t$, find the value of $b$.
Let $\frac{x+y-z}{z}=k, \frac{x-y+z}{y}=k, \frac{-x+y+z}{x}=k$.
$$
\begin{array}{l}
\left\{\begin{aligned}
x+y-z & =k z \cdots \cdots(1) \\
x-y+z & =k y \cdots \cdots(2) \\
-x+y+z & =k x \cdots \cdots(3)
\end{aligned}\right. \\
\begin{array}{l}
1)+(2)+(3): x+y+z=k(x+y+z)
\end{array} \\
\Rightarrow k=1
\end{array}
$$

From (1), $x+y=2 z,(2): x+z=2 y,(3): y+z=2 x$
$$
\begin{array}{l}
\therefore a=\frac{(x+y) \cdot(y+z) \cdot(z+x)}{x y z}=\frac{8 x y z}{x y z}=8 \\
u+t+u t=34 \Rightarrow 1+u+t+u t=35 \\
\Rightarrow(1+u)(1+t)=35 \\
\Rightarrow 1+u=5,1+t=7 \\
u=4, t=6 \\
\Rightarrow b=4+6=10
\end{array}
$$