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The medians of a triangle satisfy the triangular inequality. Moreover, if $\Delta_M$ is the area of the triangle having the medians as sides and $\Delta_T$ is the area of the original triangle, $\Delta_T=\frac{4}{3}\Delta_M$ holds, so it is sufficient to find $\Delta_M=6\sqrt{6}$ through Heron's formula in order to have $\Delta_T=8\sqrt{6}$ .

The area if the triangle = 3 * area of triangle formed out of 2/3 the median.
Area of this 2/3 midian:-
sides = 2/3(5,6,7 ). Area =(2/3)^2 * area of (5,6,7)
s=(5+6+7)/2 = 9......... s - a = 4,.......s - b = 3,..........s - c = 2
So required area = 3 * (4/9) * sqrt{9 * 4 * 3 * 2}
=8 * sqrt(6)...........= .a * sqrt(b)....>>>> a +b = 14

a=5 , b=6 , c=7

area = 1/2 x a x b = ...........1/2 x 5 x 6 = 15

15 = 5 x -/`9``` = 5+9 = 14 ......... ^_^

Refer this