374. Solve the system of equations:

$$
\left\{\begin{array}{l}
\frac{1}{x}+\frac{1}{y+z}=\frac{1}{2} \\
\frac{1}{y}+\frac{1}{z+x}=\frac{1}{3} \\
\frac{1}{z}+\frac{1}{x+y}=\frac{1}{4}
\end{array}\right.
$$
$\triangle$ In each of the equations of the system, we will eliminate the denominators of the fractions:

$$
\begin{aligned}
& 2(x+y+z)=x y+x z \\
& 3(x+y+z)=x y+y z \\
& 4(x+y+z)=x z+y z
\end{aligned}
$$

Let $x+y+z=t$. Then

$$
x y+x z=2 t, \quad x y+y z=3 t, \quad x z+y z=4 t .
$$

From this system, we will express each of the products $x y, x z$, and $y z$ in terms of $t$. The easiest way is to add all three equations:

$$
2 x y+2 x z+2 y z=9 t, \quad x y+x z+y z=\frac{9}{2} t
$$

Subtracting each of the equations of the system from the last equation, we get:

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

Multiplying all the equations of the last system:

$$
x^{2} y^{2} z^{2}=\frac{15}{8} t^{3}, \quad x y z= \pm \frac{\sqrt{30}}{4} t \sqrt{t}
$$

(Clearly, $t$ must be non-negative.)

I hope the plan for the subsequent solution is clear. Complete it independently.

Answer: $(23 / 10 ; 23 / 6 ; 23 / 2)$.