I assume that the plane on which the points lay is normal to the vector $(1,1,1)$. After a translation, we can assume that one of the points is $(0,0,0)$ and the other two have integer coordinates: let them be $(a,b,c)$ and $(d,e,f)$. Notice that the area of the triangle can be seen as the determinant of the matrix having as columns $(a,b,c)$, $(d,e,f)$ and $(1,1,1)$, divided by twice the length of $(1,1,1)$, since it is normal to the other two points (the idea is that you are calculating the volume of a solid having as base a quadrilateral having twice the area you are interested in). The determinant of a matrix with integer coefficients is integer, and the length of $(1,1,1)$ is $\sqrt{3}$. So the area is irrational.
Idea: set the origin $O$ be the circumcenter of the triangle $ABC$ and use the Pythagorean triple theorem.
Hint 1
Set the vertice $C$ on the axis $\vec{Ox'}$ (the coordinate of $C$ is $(-c,0)$ with $c\in \Bbb N^*)$. The origin $O$ is the circumcenter of the triangle $ABC$. The coordinate of points $A$ and $B$ are $(a,b)$ and $(a,-b)$ with $a,b \in \Bbb N^*$
Hint 2
The length $OA = OB = \sqrt{a^2+b^2}$ must in $\Bbb N^*$ because $OA = OB = OC$ with $C$ has integer coordinate on the axis $\vec{Ox}$
Hint 3
The solution of Pythagorean triple $a^2+b^2 = c^2$ (link) is, for example, $(a,b,c) = (m^2-n^2,2mn, m^2+n^2)$ with $m,n \in \Bbb N^*, m>n$.
Hint 4
Calculate the $\cos(\widehat{AOx}) = \frac{m^2-n^2}{m^2+n^2}$ which must be in the interval $(v_1,v_2)=\left(\cos(\frac{\pi}{3}+\epsilon), \cos(\frac{\pi}{3}-\epsilon) \right)$.
Hint 5
Prove there always exists a rational number $t\in \Bbb Q^+$ such that $$\sqrt{\frac{1+v_1}{1-v_1}} <t < \sqrt{\frac{1+v_2}{1-v_2}}$$ with $(v_1,v_2)$ defined in the hint 4. The couple $(m,n)$ will be defined by $t = \frac{m}{n}$. Hence, you deduce the value of $(a,b,c)$ from $(m,n)$
Best Answer
Hint: Assume there is a equilateral triangle whose vertices are all lattice points. Then, look at the area of the triangle using the formula $A = \dfrac{s^2\sqrt{3}}{4}$, where $s$ is the side length. Also, look at the area of the triangle using Pick's Theorem. Do you see a contradiction?