Area of the regular hexagon whose diagonal joins (2, 4) and (6, 7) is
 

We have A(2,4) and 8(6,7) as end points of diagonal of the hexagon. 

$\therefore AB=5$

So, length of the line segment joining centre of the hexagon and one of the vertices is $\frac{5}{2}$

$\therefore \text{ Area of hexagon }=6\times \text{ Area of equilateral triangle having side length }\frac{3}{2}$

$\begin{array}{r}=6\times \frac{\sqrt{3}}{4}\times \frac{25}{4}\\ =\frac{75\sqrt{3}}{8}\end{array}$