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If X,Y, Z are real and $4 x^{2} + 9 y^{2} + 16 z^{2} - 6 xy - 12 yz - 8 zx = 0$ then x, y, z are in

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A.P.

G.P.

H.P

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Correct option is C

Multiplying the given expression by 2 and rewriting it, we have(2x−3y)2+(3y−4z)2+(4z−2x)2=0⇒2x=3y=4z⇒1x,1y,1z are in A.P. ⇒x,y,z are in H.P.

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If X,Y, Z are real and 4x2+9y2+16z2−6xy−12yz−8zx=0 then x, y, z are in

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If X,Y, Z are real and $4 x^{2} + 9 y^{2} + 16 z^{2} - 6 xy - 12 yz - 8 zx = 0$ then x, y, z are in