If z1,z2,z3 are three distanct complex numbers and a,b,c are three +ve real numbers such that az2−z3=bz3−z1=cz1−z2 then a2z2−z3+b2z3−z1+c2z1−z2 is

Let az2−z3=λ⇒a2z2−z32=λ2a2z2−z3z2¯−z3¯=λ2⇒a2z2−z3=λ2z2¯−z3¯Similarly b2z3−z1=λ2z3¯−z1¯c2z1−z2=λ2z1¯−z2¯a2z2−z3+b2z3−z1+c2z1−z2=λ2z2¯−z3¯+z3¯−z1¯+z1¯−z2¯=λ20=0

If z1,z2,z3 are three distanct complex numbers and a,b,c are three +ve real numbers such that az2−z3=bz3−z1=cz1−z2 then a2z2−z3+b2z3−z1+c2z1−z2 is

Let az2−z3=λ⇒a2z2−z32=λ2a2z2−z3z2¯−z3¯=λ2⇒a2z2−z3=λ2z2¯−z3¯Similarly b2z3−z1=λ2z3¯−z1¯c2z1−z2=λ2z1¯−z2¯a2z2−z3+b2z3−z1+c2z1−z2=λ2z2¯−z3¯+z3¯−z1¯+z1¯−z2¯=λ20=0

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If $z_{1}, z_{2}, z_{3}$ are three distanct complex numbers and a,b,c are three +ve real numbers such that $\frac{a}{|z_{2} - z_{3}|} = \frac{b}{|z_{3} - z_{1}|} = \frac{c}{|z_{1} - z_{2}|}$ then $\frac{a^{2}}{z_{2} - z_{3}} + \frac{b^{2}}{z_{3} - z_{1}} + \frac{c^{2}}{z_{1} - z_{2}}$ is

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3abc

$(a b c)^{3}$

a+b+c

answer is D.

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Detailed Solution

$L e t \frac{a}{|z_{2} - z_{3}|} = λ \Rightarrow \frac{a^{2}}{{|z_{2} - z_{3}|}^{2}} = λ^{2}$

$\frac{a^{2}}{(z_{2} - z_{3}) (\bar{z_{2}} - \bar{z_{3}})} = λ^{2} \Rightarrow \frac{a^{2}}{z_{2} - z_{3}} = λ^{2} (\bar{z_{2}} - \bar{z_{3}})$

$S i m i l a r l y \frac{b^{2}}{z_{3} - z_{1}} = λ^{2} (\bar{z_{3}} - \bar{z_{1}})$

$\frac{c^{2}}{z_{1} - z_{2}} = λ^{2} (\bar{z_{1}} - \bar{z_{2}})$

$\frac{a^{2}}{z_{2} - z_{3}} + \frac{b^{2}}{z_{3} - z_{1}} + \frac{c^{2}}{z_{1} - z_{2}} = λ^{2} (\bar{z_{2}} - \bar{z_{3}} + \bar{z_{3}} - \bar{z_{1}} + \bar{z_{1}} - \bar{z_{2}})$