(a) Not exist. Suppose the contrary, then we can translate the triangle till one of the vertices coincide with the origin while the other two still have integral coordinates. Consider it on the Argand plane, let the two non-origins points as $ A=a+bi$ and $ B=c+di$, $ a,b,c,d$ are integers. Then $ B = Ae^{\frac{i\pi}{3}}$, (or vice versa) So $ (a+bi)(\frac{1}{2} +\frac{\sqrt{3}i}{2}) = (\frac{a}{2}-\frac{\sqrt{3}b}{2})+(\frac{b}{2}+\frac{\sqrt{3}a}{2})i$ So $ c=\frac{a}{2}-\frac{\sqrt{3}b}{2}$, $ d = \frac{b}{2}+\frac{\sqrt{3}a}{2}$, i.e. $ 2c-a = \sqrt{3}b$ is integer, i.e. $ b = 0$ (coz $ c,a,b$ are integers), similarly $ a = 0$. Thus the triangle becomes degenerated, which is not desirable. So not exist. (b) Consider $ (0,0,0), (7,1,0), (3,4,5)$, which forms an equilateral triangle of side length $ 5\sqrt{2}$