Let $a,b,c$ be distinct non-zero complex numbers with $\left|a\right|=\left|b\right|=\left|c\right|.$ If each of the equations $a{z}^2+bz+c=0$ and $b{z}^2+cz+a=0$ has a root having modulus 1 then which is/are correct

$z_1+z_2=-\frac{b}{a}\Rightarrow \left|z_1+z_2\right|=1$

$z_1{z}_2=\frac{c}{a}\Rightarrow \left|z_1{z}_2\right|=1$

Now $\left|z_1\right|=1\Rightarrow \left|z_2\right|=1$

$\left|z_1+z_2\right|\Rightarrow \left(z_1+z_2\right)+\left({\overline{z}}_1+{\overline{z}}_2\right)=1\Rightarrow 1+1+z_1{\overline{z}}_2+z_2{\overline{z}}_1=1\Rightarrow 1+\frac{z_1}{z_2}+\frac{z_2}{z_1}=0$

$\Rightarrow z_1^2+z_2^2-z_1{z}_2\Rightarrow {\left(z_1+z_2\right)}^2\Rightarrow {\left(-\frac{b}{a}\right)}^2=\frac{c}{a}\Rightarrow b^2=ac,c^2=ab$

$\therefore Multiply{a}^2=bc$

$\therefore Adding\Rightarrow a^2+b^2+c^2=ab+bc+ca\Rightarrow {\left(a-b\right)}^2+{\left(b-c\right)}^2+{\left(c-a\right)}^2=0$

Also ${\left(a-b\right)}^2+{\left(b-c\right)}^2+{\left(c-a\right)}^2=0$

Add & Subtract $2\left(a-b\right)\left(b-c\right)$

$\Rightarrow \left(a-b+b-c\right)+{\left(c-a\right)}^2=2\left(a-b\right)\left(b-c\right)$$\Rightarrow {\left|c-a\right|}^2=\left|a-b\right|\left|b-c\right|$

ie by symmetric

${\gamma }^2=\alpha \beta ,{\beta }^2=\gamma \alpha ,{\gamma }^2=\alpha \beta$$\Rightarrow {\alpha }^2+{\beta }^2+{\gamma }^2=\alpha \beta +\beta \gamma +\gamma \alpha$

$\Rightarrow {\left(\alpha -\beta \right)}^2+{\left(\beta -\gamma \right)}^2+{\left(\gamma -\alpha \right)}^2=0$$\Rightarrow \alpha =\beta =\gamma \Rightarrow \left|a-b\right|=\left|b-c\right|=\left|c-a\right|$