Answer: 400

Small Hint:

The intersection of the two median lines is the centroid

Big Hint:

Parametrize $A$ and $B$ from the centroid along direction vectors $(1,1)$ and $(1,2)$

Solution:

The median lines meet at the centroid $G=(-1,2).$ For real $u$ and $v,$ write $$\begin{aligned} A&=G+(u,u),\\ B&=G+(v,2v). \end{aligned}$$ Since $A+B+C=3G,$ $$C=G-(u,u)-(v,2v).$$ The condition $AC\perp BC$ gives $$\begin{aligned} &(2u+v,2u+2v)\\ &\qquad\mathbin{\cdot}(u+2v,u+4v)=0, \end{aligned}$$ or $$4u^2+15uv+10v^2=0.$$ Meanwhile $AB=60$ gives $$\begin{aligned} &(u-v)^2+(u-2v)^2\\ &\qquad=2u^2-6uv+5v^2\\ &\qquad=3600. \end{aligned}$$ Subtracting half of the first equation from the second yields $-\frac{27}{2}uv=3600,$ so $uv=-\frac{800}{3}.$

Using the two perpendicular legs from $C,$ the area is half the absolute determinant: $$\begin{aligned} [ABC] &=\frac12\left| \det\begin{pmatrix}2u+v&2u+2v\\u+2v&u+4v\end{pmatrix} \right|\\ &=\frac12|3uv| =400. \end{aligned}$$