Find area of equilateral $Delta ABC $

It's rather a comment to the Carl Schildkraut's answer, it's not in a way an answer itself. I already mentioned a link in the comments to the question for which this could be a duplicate, except this question conversely wants coordinates. Vectors are more neat here, IMHO.
Following the steps in the answer we have
0. Expand the things $$begin{cases} 4 t^2 - 4 t x + x^2 + y^2 = 100\ 4 t^2 + 2 t x - 2 sqrt{3} t y + x^2 + y^2 = 64\ 4 t^2 + 2 t x + 2 sqrt{3} t y + x^2 + y^2 = 36 end{cases}$$ 1. $$begin{cases} 4 t^2 - 4 t x + x^2 + y^2 = 100\ 4 t^2 + 2 t x - 2 sqrt{3} t y + x^2 + y^2 = 64\ 4 sqrt{3} t y = 36-64 end{cases}$$ 2. $$begin{cases} 4 t^2 - 4 t x + x^2 + y^2 = 100\ 6 t x - 2 sqrt{3} t y = 64-100\ 2 sqrt{3} t y = -14 end{cases}$$ $$begin{cases} 4 t^2 - 4 t x + x^2 + y^2 = 100\ 6 t x = 64-100-14=-50\ 2 sqrt{3} t y = -14 end{cases}$$ 3. $$begin{cases} 4 t^4 - 4 t^2 ( t x ) + t^2 x^2 + t^2 y^2 = 100t^2\ 4 t x =-frac{100}{3}\ 2 sqrt{3} t y = -14 end{cases}$$ $$begin{cases} 12 t^4 - 3 t^2cdot (4 t x ) + 3 t^2 x^2 + 3 t^2 y^2 = 300t^2\ 4 t x =-frac{100}{3}\ sqrt{3} t y = -7 end{cases}$$ $$12t^4-200t^2+frac{625}{3}+49=0$$ $$36t^4-600t^2+772=0$$ $$9t^4-150t^2+193=0$$ $$frac{D}{4}=75^2-9cdot 193=36^2cdot 3$$ $$t^2=frac{75pm 36sqrt{3}}{9}$$ $$t=pm_2sqrt{frac{25}{3}pm_1 4sqrt{3}}$$ Any questions welcome.

Orient the triangle to be centered at $(0,0)$ and let $t$ be $frac{s}{2sqrt{3}}$ so that $$A=(2t,0), B=(-t,tsqrt{3}), C=(-t,-tsqrt{3}).$$ We have some point $P=(x,y)$ so that $$(2t-x)^2+y^2=100, (t+x)^2+(y-tsqrt{3})^2=64, (t+x)^2+(y+tsqrt{3})^2=36.$$ I won't do out the algebra here because it's rather messy, but here's the general strategy:

1. Subtract the second and third equations to solve for $y$ in terms of $t$.
2. Subtract the first and second equations to solve for $x$ in terms of $t$ (once you know $y$).
3. Plug this pair $(x,y)$ into one of the three equations, giving a quadratic in $t^2$ (I think). Solve, discard unreasonable solutions, and you're done.