If z1, z2, z3 are the vertices of a triangle such that arg⁡z3−z22z1−z2−z3=π3 and 2z1−z3=z2−z3=5 then the value of z1 z¯1 1z2 z¯2 1Z3 z¯3 1 equals.

Let Az1,Bz2,Cz3 Let D be the mid point of BCBC=5⇒DC=5/2AC=5/2DC=AC,ADC equilatral ΔADC=34522=25316G.E=2(ΔABC)=4(ΔADC)=4×25316=2534

If z1, z2, z3 are the vertices of a triangle such that arg⁡z3−z22z1−z2−z3=π3 and 2z1−z3=z2−z3=5 then the value of z1 z¯1 1z2 z¯2 1Z3 z¯3 1 equals.

Let Az1,Bz2,Cz3 Let D be the mid point of BCBC=5⇒DC=5/2AC=5/2DC=AC,ADC equilatral ΔADC=34522=25316G.E=2(ΔABC)=4(ΔADC)=4×25316=2534

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If z₁, z₂, z₃ are the vertices of a triangle such that $\arg (\frac{z_{3} - z_{2}}{2 z_{1} - z_{2} - z_{3}}) = \frac{π}{3}$ and $2 |z_{1} - z_{3}| = |z_{2} - z_{3}| = 5$ then the value of $|\begin{array}{lll} z_{1} {\bar{z}}_{1} 1 \\ z_{2} {\bar{z}}_{2} 1 \\ Z_{3} {\bar{z}}_{3} 1 \end{array}|$ equals.

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$\frac{25}{4} \sqrt{3}$

$\frac{25}{8} \sqrt{3}$

$\frac{25}{2} \sqrt{3}$

$25 \sqrt{3}$

answer is A.

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Detailed Solution

$Let A (z_{1}), B (z_{2}), C (z_{3}) Let D be the mid point of BC$

$\begin{array}{l} BC = 5 \Rightarrow DC = 5 / 2 \\ AC = 5 / 2 \end{array}\} DC = AC, ADC equilatral$
$ΔADC = \frac{\sqrt{3}}{4} {(\frac{5}{2})}^{2} = \frac{25 \sqrt{3}}{16}$
$G . E = 2 (ΔABC) = 4 (ΔADC) = \frac{4 \times 25 \sqrt{3}}{16} = \frac{25 \sqrt{3}}{4}$