$A(z_1)$, $B(z_2)$ and $C(z_3)$ be vertices of ABC s/t $|z_1|=|z_2|=|z_3|=1$, $z_1+z_2\cos\alpha+z_3\sin\alpha=0$ then find $\bar z_2z_3+z_2\bar z_3$ 
If $A(z_1)$, $B(z_2)$ and $C(z_3)$ be the vertices of a triangle ABC such that $|z_1|=|z_2|=|z_3|=1$ and there exist $\alpha\in(0,\frac\pi2)$ such that $z_1+z_2\cos\alpha+z_3\sin\alpha=0$ then find the value of $\bar z_2z_3+z_2\bar z_3$ and the maximum area of triangle $ABC$.

My Attempt:
$A,B,C$ lie on the circle $|z|=1$. Thus, circumcentre is origin.
First I thought maybe the angle $\alpha$ is angle between two sides. But since that is not explicitly mentioned, I think maybe angle $\alpha$ could be just anything.
Then I thought maybe $\cos\alpha$, $\sin\alpha$ is a hint to apply inequalities. I tried but in vain.
I tried squaring the given equation but couldn't finish.
Edit:
Thanks to Hari Shankar's answer below, I now know the answer to the first part.
For area of triangle, I used determinant form and got
$$z_2\bar{z_3}(\sin\alpha+\cos\alpha+1)$$
Is this correct? How to find maximum from this?
The answer given is $1.21$.
 A: $\left|z_2 \cos \alpha + z_3 \sin \alpha \right|^2 = 1 \Rightarrow Re \left(z_2 \overline{z_3}\right) =0 \Rightarrow z_2 \overline{z_3}= \pm i$
Use the determinant form of the area to get max area as $\dfrac{\sqrt 2}{4}$
Edit: I realized I had made a mistake in the area computation
Elaborating on the area computation: The area is given by $|A|$ where
$A =\dfrac{i}{4} \begin{vmatrix}
z_1&\overline{z_1}&1\\
z_2&\overline{z_2}&1\\
z_3&\overline{z_3}&1\\
\end{vmatrix} \displaystyle \xrightarrow{\displaystyle R_1 \rightarrow R_1+R_2 \cos \alpha +R_3 \sin \alpha}\dfrac{i}{4} \begin{vmatrix}
0& 0 & 1+\cos \alpha+\sin \alpha\\
z_2  &\overline{z_2} & 1\\
z_3 &\overline{z_3} & 1\\
\end{vmatrix}$
$ = \dfrac{i}{4} \left(1+\cos \alpha + \sin \alpha\right) \ \left[z_2 \overline{z_3}-z_3 \overline{z_2} \right] = -\dfrac{\left(1+\cos \alpha + \sin \alpha\right)}{2} $
and hence $|A| \le \dfrac{1+\sqrt 2}{2} \approx 1.212$
A: As Hari Shankar observed:
$$
-z_1=z_2\cos a+z_3\sin a \quad\Longrightarrow\quad 
\mathrm{Rm}\,(z_2\overline{z}_3)=0 \quad\Longrightarrow\quad z_2\overline{z}_3=\pm i.
$$
But $z_2\overline{z}_3=\pm i$ implies that
$$
\frac{z_2}{z_3}=\frac{z_2\overline{z}_3}{|z_3|^2}=\pm i.
$$
Hence $z_2=\pm i z_3$. Nothing changes in the problem if we rotate the $z_j$'s all by the same angle, i.e., nothing changes is $z_1,z_2,z_3$ are replaced by $e^{it}z_1,e^{it}z_2,e^{it}z_3$. So without loss of generality we assume that $z_2=1$, $z_3=\pm i$, and
$$
z_1=-(z_2\cos a+z_3\sin a)=-(\cos a \pm i\sin a)=e^{i(\pi\pm a)}.
$$
Clearly, one of the sides of the triangle, the $z_2z_3$, has length $\sqrt{2}$, in which case the inscribed in the unit circle triangle of maximum area is the isosceles of basis $\sqrt{2}$ and height $1+\sqrt{2}/2$, and hence area $A=\frac{1}{2}(1+\sqrt{2})$.
