Let the points of intersections of the lines $$x - y + 1 = 0$$, $$x - 2y + 3 = 0$$ and $$2x - 5y + 11 = 0$$ are the mid points of the sides of a triangle ABC. Then the area of the triangle ABC is _________.

We have the three straight lines $$$x - y + 1 = 0\;,\; x - 2y + 3 = 0\;,\; 2x - 5y + 11 = 0$$$. Their pair-wise points of intersection will give us three points, and the question tells us that these very points are the mid-points of the sides of a triangle $$ABC$$.

First we find the intersection of $$x - y + 1 = 0$$ and $$x - 2y + 3 = 0$$. From $$x - y + 1 = 0$$ we get $$y = x + 1$$. Substituting this in $$x - 2y + 3 = 0$$, we obtain $$$x - 2(x + 1) + 3 = 0 \;\Longrightarrow\; x - 2x - 2 + 3 = 0 \;\Longrightarrow\; -x + 1 = 0 \;\Longrightarrow\; x = 1.$$$ Putting $$x = 1$$ back in $$y = x + 1$$ gives $$y = 2$$. So the first point is $$M_1(1,\,2).$$

Next we intersect $$x - 2y + 3 = 0$$ with $$2x - 5y + 11 = 0$$. We keep the first equation as it is and multiply it by $$2$$ to get a convenient pair: $$$2(x - 2y + 3) = 0 \;\Longrightarrow\; 2x - 4y + 6 = 0.$$$ Now we subtract this from $$2x - 5y + 11 = 0$$: $$$(2x - 5y + 11) - (2x - 4y + 6) = 0 \;\Longrightarrow\; -y + 5 = 0 \;\Longrightarrow\; y = 5.$$$ Putting $$y = 5$$ in $$x - 2y + 3 = 0$$ gives $$$x - 2(5) + 3 = 0 \;\Longrightarrow\; x - 10 + 3 = 0 \;\Longrightarrow\; x = 7.$$$ Hence the second point is $$M_2(7,\,5).$$

Finally, we intersect $$2x - 5y + 11 = 0$$ with $$x - y + 1 = 0$$. Again, from $$x - y + 1 = 0$$ we have $$y = x + 1$$. Substituting in $$2x - 5y + 11 = 0$$, we get $$$2x - 5(x + 1) + 11 = 0 \;\Longrightarrow\; 2x - 5x - 5 + 11 = 0 \;\Longrightarrow\; -3x + 6 = 0 \;\Longrightarrow\; x = 2.$$$ Putting $$x = 2$$ in $$y = x + 1$$ yields $$y = 3.$$ So the third point is $$M_3(2,\,3).$$

Thus the mid-points of $$\triangle ABC$$ are $$$M_1(1,\,2),\; M_2(7,\,5),\; M_3(2,\,3).$$$

Now, for any triangle, the triangle formed by the three mid-points (called the medial triangle) has exactly one-fourth the area of the original triangle. Hence $$$\text{Area of } \triangle ABC = 4 \times \text{Area of } \triangle M_1M_2M_3.$$$ So we must first compute the area of $$$\triangle M_1M_2M_3.$

The coordinate-geometry formula for the area of a triangle with vertices $$$(x_1,y_1),\,(x_2,y_2),\,(x_3,y_3)$$ is $$\text{Area} = \frac12\;\bigl|\,x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\bigr|.$$ Taking $$(x_1,y_1) = (1,2),\; (x_2,y_2) = (7,5),\; (x_3,y_3) = (2,3)$$$, we substitute step by step:

$$$\begin{aligned} \text{Area}(M_1M_2M_3) &= \frac12\;\Bigl|\, 1\,(5 - 3) + 7\,(3 - 2) + 2\,(2 - 5) \Bigr| \\ &= \frac12\;\Bigl|\, 1 \times 2 \;+\; 7 \times 1 \;+\; 2 \times (-3) \Bigr| \\ &= \frac12\;\bigl|\,2 + 7 - 6\bigr| \\ &= \frac12\;\bigl|\,3\bigr| \\ &= \frac{3}{2}. \end{aligned}$$

Therefore $$\text{Area of } \triangle ABC = 4 \times \frac{3}{2} = 6.$$

So, the answer is $$6$$.