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 $\mathcal{A}$ be the area of $\triangle ABC$.
Let $\theta$ and $\phi$ be the angles $\angle PAC$ and $\angle BAP$ respectively.
We have $\theta + \phi = \angle BAC = \frac{\pi}{3}$.
As functions of $\theta$ and $\phi$, the side lengths $b$, $c$ and area $\mathcal{A}$ are:
$$
\begin{cases}
c(\theta) &= \cos\theta + \sqrt{2^2-\sin^2\theta}\\
b(\phi) &= \cos\phi + \sqrt{3^2-\sin^2\phi}
\end{cases}
\quad\text{ and }\quad
\mathcal{A}(\theta) = \frac{\sqrt{3}}{4} c(\theta)b\left(\frac{\pi}{3}-\theta\right)
$$
In order for $\mathcal{A}(\theta)$ to achieve maximum has a particular $\theta$,
we need
$$\frac{d\mathcal{A}}{d\theta} = 0
\iff \frac{1}{\mathcal{A}}\frac{d\mathcal{A}}{d\theta} = 0
\iff \frac{1}{c}\frac{dc}{d\theta} - \frac{1}{b}\frac{db}{d\phi} = 0
\iff \frac{\sin\theta}{\sqrt{2^2-\sin^2\theta}} - \frac{\sin\phi}{\sqrt{3^2-\sin^2\phi}} = 0$$
This implies
$$\frac{\sin\theta}{2} = \frac{\sin\phi}{3}
 = \frac13 \sin\left(\frac{\pi}{3} - \theta\right)
 = \frac13 \left(\frac{\sqrt{3}}{2}\cos\theta - \frac12\sin\theta\right)
\iff 4\sin\theta = \sqrt{3}\cos\theta$$
and hence
$$\theta = \tan^{-1}\left(\frac{\sqrt{3}}{4}\right) \approx 0.4086378550975924
\;\;( \approx 23.41322444637054^\circ )$$
Furthermore, we have
$\displaystyle\;\frac{\sin\theta}{2} = \frac{\sin\phi}{3} = \frac{\sqrt{3}}{2\sqrt{19}}\;$.
Substitute this into the expression for side lengths and area, we get
$$
\begin{cases}
c &= \frac{4+\sqrt{73}}{\sqrt{19}}\\
b &= \frac{7+3\sqrt{73}}{2\sqrt{19}}
\end{cases}
\quad\implies\quad
\mathcal{A} = \frac{\sqrt{3}}{8}(13+\sqrt{73}) \approx 4.664413635668018
$$
Please note that the condition $\displaystyle\;\frac{\sin\theta}{2} = \frac{\sin\phi}{3}$ is equivalent to $\angle ABP = \angle ACP$. If one can figure out why these two angles equal to each other when $\mathcal{A}$ is maximized, one should be able to derive all the result here w/o using any calculus.
