Let  point of intersection of lines $x-y+1=0$  and  $x-2y+3=0$ and  $2x-5y+11=0$  are midpoints of sides of triangle ABC. Then Area of triangle ABC is 

Solving system of equations we get point of intersections $(1,2)(7,5)and(2,3)$  

Area of   $\Delta DEF=\frac{1}{2}\left|\begin{array}{ccc}1& 2& 1\\ 7& 5& 1\\ 2& 3& 1\end{array}\right|$
$\begin{array}{l}\Rightarrow \frac{1}{2}|1(5-3)-2(7-2)+1(21-10)|\\ \Rightarrow \frac{1}{2}|2-10+11|\Rightarrow \frac{3}{2}\\ 4areaof\Delta DEF=areaof\Delta ABC\Rightarrow 4\left(\frac{3}{2}\right)=6sq.units\end{array}$