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Question Number 221393 by universe last updated on 02/Jun/25
     a, b, c are complex number and      ∣a∣ = ∣b∣=∣c∣= 1 and  (a^2 /(bc))+(b^2 /(ac)) +(c^2 /(ab)) = −1      where ∣.∣ is  modules function  then ∣a+b+c∣ can be   (A) 0       (B) 1           (C) (3/2)          (D) 2
$$\:\:\:\:\:{a},\:{b},\:{c}\:{are}\:{complex}\:{number}\:{and}\: \\ $$$$\:\:\:\mid{a}\mid\:=\:\mid{b}\mid=\mid{c}\mid=\:\mathrm{1}\:{and}\:\:\frac{{a}^{\mathrm{2}} }{{bc}}+\frac{{b}^{\mathrm{2}} }{{ac}}\:+\frac{{c}^{\mathrm{2}} }{{ab}}\:=\:−\mathrm{1} \\ $$$$\:\:\:\:{where}\:\mid.\mid\:{is}\:\:{modules}\:{function} \\ $$$${then}\:\mid{a}+{b}+{c}\mid\:{can}\:{be}\: \\ $$$$\left({A}\right)\:\mathrm{0}\:\:\:\:\:\:\:\left({B}\right)\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\left({C}\right)\:\frac{\mathrm{3}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\left({D}\right)\:\mathrm{2} \\ $$

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