Let's calculate the value of $(x+y+z)^{2}$, if

$$
2 x(y+z)=1+y z, \quad \frac{1}{x}-\frac{2}{y}=\frac{3}{2} \quad \text { and } \quad x+y+\frac{1}{2}=0 .
$$
I. Solution: None of the variables can be 0, so the fractions can be eliminated, and we obtain the following system of equations:

$$
2 x y+2 x z-y z=1, \quad 3 x y+2 x z-y z=0, \quad x z+y z=-1 .
$$

This is a first-degree system of equations if we consider $x y, x z, y z$ as unknowns. Subtracting the second equation from the first, and then multiplying the consecutive equations by 3, -2, 1, and finally by 3, -2, -2, we get in sequence that

$$
x y=-1, \quad x z=\frac{2}{3}, \quad y z=-\frac{5}{3}
$$

From here,

$$
x^{2}=\frac{x y \cdot x z}{y z}=\frac{2}{5} \quad y^{2}=\frac{x y \cdot x y}{x z}=\frac{5}{2}, \quad z^{2}=\frac{x z \cdot y z}{x y}=\frac{10}{9}
$$

Using these,

$$
\begin{gathered}
(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2(x y+x z+y z)= \\
=\frac{2}{5}+\frac{5}{2}+\frac{10}{9}+2\left(-1+\frac{2}{3}-\frac{5}{3}\right)=\frac{1}{90}
\end{gathered}
$$

II. Solution: From the third equation, substituting the value of $x+y$,

$$
(x+y+z)^{2}=\left(z-\frac{1}{z}\right)^{2}=z^{2}+\frac{1}{z^{2}}-2
$$

so it is enough to determine the value of $z^{2}$. Using the form of the equations as in the previous solution, add the first and last equations:

$$
2 x y+3 x z=0, \quad y=-\frac{3}{2} z
$$

Substituting this into the second equation,

$$
-\frac{9}{2} x z+2 x z+\frac{3}{2} z^{2}=0, \quad \text { hence } \quad x=\frac{3}{5} z
$$

Substituting the obtained values into the first equation,

$$
-\frac{9}{5} z^{2}+\frac{6}{5} z^{2}+\frac{3}{2} z^{2}=1, \quad z^{2}=\frac{10}{9}
$$

thus,

$$
(x+y+z)^{2}=\frac{10}{9}+\frac{9}{10}-2=\frac{1}{90}
$$