falzoon's folly ....

If the sides are $n-1, n, n+1$ , then the area is $\frac{n}4 \sqrt{3(n^2-4)}$ by Heron's formula. Clearly $n$ must be even, say $n = 2a$ . Then we get an area of $a\sqrt{3(a^2-1)}$ . So $a^2-1=3b^2$ for some integer $b$ and the area is $3ab$ . The first three solutions of the Pell equation $a^2-3b^2 = 1$ are $(a,b) = (2,1), (7,4), (26,15)$ , which leads to $n = 4, 14, 52, \ldots$ . The third triangle has area $3 \cdot 26 \cdot 15 = \fbox{1170}$ .