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If x,y,z satisfy given system of equations: `x^2+y^2+z^2=133`,`y+z-x=7` and `yz=x^2` , then the value of x+y+z is a. 12 b. 15 c. 19 d. 0

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`x^2+y^2+z^2 = 133`
`=>x^2+y^2+z^2+2yz-2yz = 133`
`=>x^2-2yz+(y+z)^2 = 133->(1)`
As, `y+z-x = 7 => y+z = 7+x`
So, putting value of `y+z` and `yz` in (1),
`=>x^2-2x^2+(7+x)^2 = 133`
`=> -x^2+49+x^2+14x = 133`
`=>14x = 84=> x = 6`
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