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If x + y + z = 19,x^2 +y^2+z^2=133 and x...

If x + y + z = 19,`x^2 +y^2+z^2=133` and `xz=y^2, x>z>0`, what is the value of (x-z) ?
यदि x + y + z = 19,`x^2 +y^2+z^2=133` और `xz=y^2, x>z>0` तो (x-z) का मान ज्ञात करें:

A

`-2`

B

0

C

`-5`

D

5

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