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 E be the point outside triangle such that AEPC is a parallelogram , let F be point outside the triangle such that CPBF is a parallelogram, and let D be a point outside ABC such that APBC is paralellogram. Now, you have hexagon AECFBD with sides EC=1,CF=2,FB=3,BD=1, AD=2, EA=3, and draw line EB. Then, ADBE and EBFC are the same(congruent) and now surface of a hexagon is double of a surface of ADEB. Now, in that, you have three sides which are determined, plus AD is parallel to EB. Now consider ADBE. There, ED=1,DA=2,AB=3. Let M be perpendicular to EB from D and let N be perpendicular to EB from A. Now let EM be x, and NB be y. Let DM be h, as well. We have to maximise the surface of ADEB. We can see that surface is sum of surfaces of NMDA and EMD and ANB. So, it is $2h+\frac{hx}{2}+\frac{hy}{2}$. Relations that we have imply from triangles EMD and ANB, Now, those are $1=h^2+x^2, 9 = h^2+y^2$ so $y^2-x^2=8$ ,and $h^2=1-x^2$.
