Question

If an equilateral triangle $ABC$ with vertices at $z_1,z_2$ and $z_3$ be inscribed in a circle $|z|=2$ and again a circle is inscribed in $△ABC$ touching sides $AB,BC$ and $CA$ at $D\left(z_4\right),E\left(z_5\right)$ and $F\left(z_6\right)$ respectively.

Column I		Column II
P	Value of $\mathrm{Re}\left(z_1{\overline{z}}_2+z_2{\overline{z}}_3+z_3{\overline{z}}_1\right)$ is equal to	1	2
Q	If $\frac{4z_1}{z_3}=a(-1+i\sqrt{3})$ then a is equal to	2	-6
R	${\left|z_1+z_2\right|}^2+{\left|z_2+z_3\right|}^2+{\left|z_3+z_1\right|}^2$ is equal to	3	12
S	If $P$ is any point on incircle, then $D{P}^2+E{P}^2+F{P}^2$ is	4	6

• A. P= 1 ,Q=3 ,R= 4 ,S= 3
• B. P=3 ,Q=4 ,R=1 ,S= 2
• C. P= 2 ,Q=1 ,R=3 ,S= 4
• D. P=2 ,Q=3,R= 2 ,S= 1

Answer 1

Answer: (C)

(A)
$\Theta {\left|z_1-z_2\right|}^2={\left|z_1\right|}^2+{\left|z_2\right|}^2-2\mathrm{Re}\left(z_1{\bar{z}}_2\right)$
$\Rightarrow (2\sqrt{3}{)}^2=2^2+2^2-2\mathrm{Re}\left(z_1{\bar{z}}_2\right)$
$\Rightarrow \mathrm{Re}\left(z_1{\bar{z}}_2\right)=-2$
Similarly, $\mathrm{Re}\left(z_2{\overline{z}}_3\right)=-2$
& $\mathrm{Re}\left(z_3{\overline{z}}_1\right)=-2$
$\therefore \mathrm{Re}\left(z_1{\overline{z}}_2+z_2{\overline{z}}_3+z_3{\overline{z}}_1\right)=-6$
(B) $\Theta \frac{z_1-0}{z_3-0}=e^{\frac{i2\pi }{3}}\Rightarrow \frac{z_1}{z_3}=\left(-\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)\Rightarrow \frac{4z_1}{z_3}=2(-1+i\sqrt{3})$
$\therefore a=2$
(C)${\left|z_1+z_2\right|}^2+{\left|z_2+z_3\right|}^2+{\left|z_3+z_1\right|}^2=2\left({\left|z_1\right|}^2+{\left|z_2\right|}^2+{\left|z_3\right|}^2\right)+2\left(\mathrm{Re}\left(z_1{\bar{z}}_2+z_2{\bar{z}}_3+z_3{\bar{z}}_1\right)\right)$
$=2(4+4+4)+2(-6)=12$
(D) $D{P}^2+E{P}^2+F{P}^2={\left|z-z_4\right|}^2+{\left|z-z_5\right|}^2+{\left|z-z_6\right|}^2=3|z{|}^2+{\left|z_4\right|}^2+{\left|z_5\right|}^2+{\left|z_6\right|}^2=6$