Since $x, y, z$ are greater than $3$, it follows that $y+z-2, z+x-4, x+y-6$ are positive. Thus, from Cauchy-Schwarz inequality we get:
$$\left( \frac{(x+2)^2}{y+z-2} + \frac{(y+4)^2}{x+z-4} + \frac{(z+6)^2}{x+y-6} \right) \left( (y+z-2) + (x+z-4) + (x+y-6) \right) \geq (x+y+z+12)^2 \\ \Leftrightarrow \frac{(x+2)^2}{y+z-2} + \frac{(y+4)^2}{x+z-4} + \frac{(z+6)^2}{x+y-6} \geq \frac{1}{2} \cdot \frac{(x+y+z+12)^2}{(x+y+z-6)}.$$
From the hypothesis of the problem it follows that
$$\frac{(x+y+z+12)^2}{(x+y+z-6)} \leq 72, \qquad (1)$$
where as the equality holds when:
$$\frac{x+2}{y+z-2} = \frac{y+4}{x+z-4} = \frac{z+6}{x+y-6} = \lambda \Leftrightarrow \begin{cases} \lambda(y+z) - x = 2(\lambda+1) \\ \lambda(x+z) - y = 4(\lambda+1) \\ \lambda(x+y) - z = 6(\lambda+1) \end{cases}. \qquad (2)$$
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Moreover, we observe that:
$$\frac{(x+y+z+12)^2}{(x+y+z-6)} = \frac{(x+y+z+12)^2}{(x+y+z+12)-18} = \frac{\omega^2}{\omega-18},$$
where we have put $\omega = x + y + z + 12$. Since we have
$$\frac{\omega^2}{\omega-18} \geq 4 \cdot 18 = 72 \Leftrightarrow \omega^2 - 4 \cdot 18\omega + 4 \cdot 18^2 \geq 0 \Leftrightarrow (\omega - 36)^2 \geq 0,$$
it follows that
$$\frac{(x+y+z+12)^2}{(x+y+z-6)} \geq 72. \qquad (3)$$
Equality holds when:
$$\omega = x + y + z + 12 = 36 \Leftrightarrow x + y + z = 24 \qquad (4)$$
From relations (1) and (3) follows that:
$$\frac{(x+y+z+12)^2}{(x+y+z-6)} = 72,$$
and thus equations (2) are (4) are valid, and hence we have the system
$$\left\{ \begin{array}{l} (2\lambda - 1)(x + y + z) = 12(\lambda + 1) \\ x + y + z = 24 \end{array} \right\} \Rightarrow \lambda = 1.$$
For $\lambda = 1$, from relations (2) we get the system :
$$\left\{ \begin{array}{l} y+z-x=4 \\ x+z-y=8 \\ x+y-z=12 \end{array} \right\} \Leftrightarrow (x,y,z)=(10,8,6).$$
Therefore the unique solution of the problem is the triple $(x, y, z) = (10, 8, 6)$, taking in mind that it satisfies the equation $x + y + z = 24$.