The answer is 1.

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$\begin{cases}x^2-4y+7=0 \\ y^2-6z+14=0\\ z^2-2x-7=0 \end{cases}$ Adding the three equations, we have, $x^2 - 4y + 7 + y^2 - 6z + 14 + z^2 - 2x - 7 = 0$ Rearranging the terms and completing the square, $(x-1)^2+(y-2)^2+(z-3)^2=0$ Hence $x=1, y=2, z=3$ are the only solutions and can easily be verified.

Therefore, the sum of all possible values for $x$ is $\boxed{1}$