Given that there is only one triplet \((x,y,z),\) where \(x,y\) and \(z\) are real numbers greater than \(3,\) which satisfies the equation

\[\frac{(x+2)^2}{y+z-2}+\frac{(y+4)^2}{z+x-4}+\frac{(z+6)^2}{x+y-6}=36,\]

find the value of \(x.\)

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