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If \(\dfrac{x^2 + y^2 + z^2 - 64}{xy - yz - zx} = -2\) and \(x + y = 3z\), the value of \(z\) is :

The correct choice is: 4 Cross-multiplying: \(x^2+y^2+z^2-64=-2xy+2yz+2zx\), so \(x^2+y^2+z^2+2xy-2yz-2zx=64\), i.e. \((x+y-z)^2=64\). With \(x+y=3z\):…

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If x2+y2+z2−64xy−yz−zx=−2\dfrac{x^2 + y^2 + z^2 - 64}{xy - yz - zx} = -2 and x+y=3zx + y = 3z, the value of zz is :

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Answer and explanation

Correct answer C

The correct choice is: 4

Cross-multiplying: x2+y2+z2−64=−2xy+2yz+2zxx^2+y^2+z^2-64=-2xy+2yz+2zx, so x2+y2+z2+2xy−2yz−2zx=64x^2+y^2+z^2+2xy-2yz-2zx=64, i.e. (x+y−z)2=64(x+y-z)^2=64.

With x+y=3zx+y=3z: (3z−z)2=4z2=64⇒z=4(3z-z)^2=4z^2=64\Rightarrow z=4 (taking the positive value).

Hence the answer is ‘4’.

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  • NCHM Joint Entrance Examination (NCHM JEE)

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