Is it surely Heron?

By the cosine law, we have

$338=61+153-2(\sqrt{61})(\sqrt{153})(\cos A)$

$A=\cos^{-1}\left(\dfrac{87}{-\sqrt{9333}}\right)$

So the area is

$\dfrac{1}{2}\cdot \sqrt{61} \cdot \sqrt{153} \cdot \sin \left[\cos^{-1}\left(\dfrac{87}{-\sqrt{9333}}\right)\right] = 21$

Using Heron's formula, Area=1/4*sqrt(4ab-(a+b-c)^2) Where a,b, and c are the squares of the side lengths of the triangle. It comes out to 21

first find all sides of the triangle.
after that using hero's formulae area=square root of(s(s-a)(s-b)(s-c)) where s=(a+b+c)/2 and a,b,c are the sides resp.

(sqrt(388)+sqrt(61)+sqrt(153))/2=S A=sqrt(S(S-sqrt(388))(S-sqrt(61))(S-sqrt(153)))=21