Question 3
Find all triples of real numbers $(x, y, z)$ which are greater than $3$ and satisfy the equality:
$$
\frac{(x+2)^2}{y+z-2} + \frac{(y+4)^2}{z+x-4} + \frac{(z+6)^2}{x+y-6} = 36.
$$
$$
\frac{(x+2)^2}{y+z-2} + \frac{(y+4)^2}{z+x-4} + \frac{(z+6)^2}{x+y-6} = 36.
$$