Given quadratic equation,
$${Z^2} + aZ + b = 0$$
and two roots are $${Z_1}$$ and $${Z_2}$$.

$$\therefore$$ $${Z_1}$$ + $${Z_2}$$ = $$-a$$ and $${Z_1}$$$${Z_2}$$ = $$b$$

Question says,
There are three complex numbers:
1. Origin (0)
2. $${Z_1}$$
3. $${Z_2}$$
and they form an equilateral triangle. So They are the vertices of the triangle.

[ Important Point : If $${Z_1}$$, $${Z_2}$$ and $${Z_3}$$ are the vertices of an equilateral triangle then -
$$Z_1^2$$ + $$Z_2^2$$ + $$Z_3^2$$ = $${Z_1}{Z_2}$$ + $${Z_2}{Z_3}$$ + $${Z_3}{Z_1}$$ ]

In this question,
$${Z_1}$$ = 0, $${Z_2}$$ = $${Z_1}$$ and $${Z_3}$$ = $${Z_2}$$

By putting those values in the equation we get,

$${0^2}$$ + $$Z_1^2$$ + $$Z_2^2$$ = $$0$$ + $${Z_1}{Z_2}$$ + 0

$$ \Rightarrow $$ $$Z_1^2$$ + $$Z_2^2$$ = $${Z_1}{Z_2}$$

$$ \Rightarrow $$ $$Z_1^2$$ + $$Z_2^2$$ = $$b$$ [ as $${Z_1}$$$${Z_2}$$ = $$b$$ ]

$$ \Rightarrow $$ $${\left( {{Z_1} + {Z_2}} \right)^2}$$ - $$2{Z_1}{Z_2}$$ = $$b$$

$$ \Rightarrow $$ $${\left( {{Z_1} + {Z_2}} \right)^2}$$ - $$2b$$ = $$b$$

$$ \Rightarrow $$ $${\left( {{Z_1} + {Z_2}} \right)^2}$$ = $$3b$$

$$ \Rightarrow $$ $${\left( { - a} \right)^2}$$ = $$3b$$

$$ \Rightarrow $$ $${a^2}$$ = $$3b$$

So Option (D) is correct.

[ Note : This question is asked to check if you know the following formula -

"If $${Z_1}$$, $${Z_2}$$ and $${Z_3}$$ are the vertices of an equilateral triangle then -
$$Z_1^2$$ + $$Z_2^2$$ + $$Z_3^2$$ = $${Z_1}{Z_2}$$ + $${Z_2}{Z_3}$$ + $${Z_3}{Z_1}$$" ]