6. (8 points) Let for positive numbers $x, y, z$ the system of equations holds:

$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=108 \\
y^{2}+y z+z^{2}=16 \\
z^{2}+x z+x^{2}=124
\end{array}\right.
$$

Find the value of the expression $x y+y z+x z$.
Answer: 48

Solution: Let there be three rays with vertex $O$, forming angles of $120^{\circ}$ with each other. On these rays, we lay off segments $O A=x, O B=y, O C=z$. Then, by the cosine theorem, $A B^{2}=108$, $B C^{2}=16, A C^{2}=124$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of the areas of triangles $A O B, B O C, A O C$ is equal to the area of triangle $A B C$. This gives us the relation $\frac{1}{2}(x y+y z+x z) \sin 120^{\circ}=\frac{1}{2} \cdot 6 \sqrt{3} \cdot 4$. From this, we obtain the answer.