We have:

\(x^2=y+5\)

\(y=z-2\)

\(z=2x\)

\(z=2x\)

\(y=2x-2\)

\(x^2=2x+3\)

\(x^2-2x-3=0\)

\(x=3\) or \(x=-1\)

If \(x=3\), then \(y=2x-2=4\) and \(z=2x=6\)

Therefore \(x^3+y^2+z=27+16+6=49\)

If \(x=-1\), then \(y=2x-2=-4\) and \(z=2x=-2\)

Therefore \(x^3+y^2+z=-3+16-2=13\)

If (1) is true, then the only possible x is 3, and the result is divisible by 7.

If (2) is true, then the only possible x is 3, and the result is again divisible by 7.

So, the answer is (D)

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