find the maximum possible area of $\triangle{ABC}$ Let $ABC$ be of triangle with $\angle BAC = 60^\circ$
. Let $P$ be a point in its interior so that $PA=1, PB=2$ and
$PC=3$. Find the maximum area of triangle $ABC$.
I took reflection of point $P$ about the three sides of triangle and joined them to vertices of triangle. Thus I got a hexagon having area double of triangle, having one angle $120$ and sides $1,1,2,2,3,3$. We have to maximize area of this hexagon. For that, I used some trigonometry but it went very complicated and I couldn't get the solution.

Let $\theta=\measuredangle PAB$ in the triangle you specify. Then 
$\measuredangle PAC=60°-\theta$.

By the law of cosines,
$$2^2=1^2+c^2-2\cdot 1\cdot c\cdot \cos\theta$$
$$3^2=1^2+b^2-2\cdot 1\cdot b\cdot \cos(60°-\theta)$$
Solving those equations for $b$ and $c$,
$$c=\cos\theta+\sqrt{\cos^2\theta+3}$$
$$b=\cos(60°-\theta)+\sqrt{\cos^2(60°-\theta)+8}$$
Since the triangle's area is $\frac 12bc\sin A$ and $A=60°$, putting them all together we get
$$Area=\frac{\sqrt 3}{4}\left(\cos(60°-\theta)+\sqrt{\cos^2(60°-\theta)+8}\right)\left(\cos\theta+\sqrt{\cos^2\theta+3}\right)$$
where $0°<\theta<60°$. This looks like a bear to maximize analytically, so I did it numerically with a graph and got

$$Area_{max}\approx 4.66441363567 \quad\text{at}\quad \theta\approx 23.413°$$

I could get only five significant digits for $\theta$ but the maximum area should be accurate to all shown digits. I checked this with a construction in Geogebra and it checks. WolframAlpha timed out while trying to find an exact maximum value of the triangle's area from that my formula.
