Area of $ABCD$ where $A,B,C$ lie on circle with center $D$ Let $\mathrm{A}\left(z_{1}\right), \mathrm{B}\left(z_{2}\right)$ and $\mathrm{C}\left(z_{3}\right)$ lie on the circle $|z-\mathrm{i}|=1$ and satisfy the equation $3 \mathrm{z}_{1}+\mathrm{i}=2 \mathrm{z}_{2}+2 \mathrm{z}_{3}$. If $\mathrm{D}$ is the centre of the circle $|z-\mathrm{i}|=1$, then area of quadrilateral $\mathrm{ABCD}$ equals---
I tried to analyze by using geometry but am not able to reach to final answer. Please help,
 A: If we shift the circle down by $1$ unit, we get $D=0$,
$$|z|=1$$
and $A(z_1),B(z_2),C(z_3)$ satisfy
$$3(z_1+i)+i=2(z_2+i)+2(z_3+i)$$
$$\Rightarrow 3z_1=2z_2+2z_3 \tag{1}$$
Since the expression is symmetric in $z_2,z_3$, $B,C$ lie symmetrically wrt to $DA$ (see remark below). The quadrilateral has diagonals $AD$ and $BC$ where radius $AD$ is perpendicular to chord $BC$. Its area is then half of product of diagonals.
Let arguments of $A,B,C$ be $\theta_1,\theta_2,\theta_3$. Then $BC$ subtends $2\theta=|\theta_2-\theta_3|$ at the center. So $BC=2\sin \theta$
From $(1)$, we have
$$3\cos \theta_1=2\cos \theta_2 + 2\cos \theta_3 \tag{2}$$
$$3\sin \theta_1=2\sin \theta_2 + 2\sin \theta_3 \tag{3}$$
Squaring and adding we get
$$\cos (\theta_2 - \theta_3)=\frac{1}{8}=\cos 2\theta$$
As pointed by @user, the above can also be obtained by multiplying $(1)$ by its complex conjugate equation.
Now $$\text{Area} = \frac{1}{2} \cdot DA \cdot BC = \sin \theta$$
$$=\sqrt{\frac{1-\cos 2\theta}{2}}$$
$$=\boxed{\frac{\sqrt{7}}{4}}$$
Remark :
If this is not convincing, note that $(1)$ can be rewritten as $$5z_1=2(z_1+z_2+z_3)$$
where term in brackets is coordinates of orthocenter of $\triangle ABC$. It says $A$ is collinear with orthocenter and $D$, the circumcenter $\Rightarrow \triangle ABC$ is isosceles with $AB=AC$ i.e., $B,C$ lie symmetrically on either side of $DA$.
