For the record, I will solve the much easier problem of finding infinitely many rational points on the circumcircle. Let's consider the case of a point $P$ on the arc from $A$ to $B$. By Ptolemy's theorem, $|PA| + |PB| = |PC|$, so if $|PA|$ and $|PB|$ are rational then so is $|PC|$. Also, $P$ is on the arc from $A$ to $B$ if and only if $P$ is on the right side of $\overline{AB}$ and $\angle APB = 120^{\circ}$. By the Law of Cosines, $\angle APB = 120^{\circ}$ is equivalent to
$$|PA|^2 + |PA| |PB| +|PB|^2=1.$$
This conic can be paramterized in the usual way: Put $|PA|=1+t$, $|PB|=kt$. Solve for $t$ in terms of $k$; the result is $t=-(k+2)/(k^2+k+1)$. So
$$|PA| = \frac{k^2-1}{k^2+k+1} \quad |PB| = \frac{-k^2-2k}{k^2+k+1} \quad |PC|=\frac{-2k-1}{k^2+k+1}$$
or, in other words,
$$a=k^2-1 \quad b= -k^2-2k \quad c=-2k-1 \quad d=k^2+k+1.$$
We want to have $-2 < k < -1$ to get the right signs.
I don't want the bounty for this though; I want mathlove to explain how the heck he or she found his or her solution.
I have figured out a way to get mathlove's answer. I'll write $PA=a/d$, $PB=b/d$ and $PC=c/d$. As described here, these obey the relation
$$a^4+b^4+c^4+d^4 = a^2 b^2 + a^2 c^2 + a^2 d^2 + b^2 c^2 + b^2 d^2 + c^2 d^2.$$
Let $\Sigma$ be the surface in $\mathbb{P}^3$ cut out by this degree $4$ equation. Notice that $\Sigma$ has $16$ singular points: The $4$ points $(\pm 1 : \pm 1 : \pm 1 :
0)$ and the other $12$ which come from putting the zero in the other possible positions.
The three points $(1:1:0:1)$, $(1:0:1:1)$ and $(0:1:1:1)$ are the vertices of the triangle; the other $13$ singularities involve negative or infinite values for $(PA, PB, PC)$.
The technical term for this is a Kummer surface and, in fact, $\Sigma$ is the special kind of Kummer surface called a tetrahedroid. But we don't need to know this to follow the rest of the argument.
Take a plane through any three of the singularities. The resulting planar slice of $\Sigma$ will be a degree $4$ plane curve with $\geq 3$ nodes. If there are exactly $3$ nodes, the resulting curve is genus $0$, and thus has a rational parametrization over $\mathbb{C}$. That parmetrization doesn't have to have rational coefficients but, in some lucky cases, it does. We also aren't promised that the resulting values of $(PA, PB,PC)$ will be positive, let alone inside the triangle but, again, sometimes we get luck.
Mathlove uses the plane $b+d=2a$, passing through the points $(0,1,1,-1)$, $(0,-1,1,1)$ and $(1,1,0,1)$. The complete list of planes, up to permuting $(a,b,c,d)$ and switching signs, is $a=0$, $a+b=0$, $a+b+c=0$, $a+b+2c=0$ and $a+b+2c+3d=0$. These symmetries of $\Sigma$ do not respect the condition that the points actually correspond to physical points inside the triangle, so you have to keep track of more possibilities if you want that to hold.
Several of these planes correspond to interesting geometric configurations:
The equations $a+d=b$, $a+b=d$ and $b+d=a$ are $PA+1=PB$, $PA+PB=1$ and $PB+1=PA$ respectively, which say that $P$ lies on the line $AB$ (either in the two unbounded rays or in the line segment $AB$.)
The equation $a+b=c$ means $PA+PB=PC$, so (by Ptolemy's theorem) $P$ is on the arc of the circumcircle from $A$ to $B$; the other arcs of the circumcircle are described similarly.
The equation $a=b$ means that $P$ is on the perpendicular bisector of $AB$. This doesn't actually contribute any points; the intersection of $\Sigma$ with $\{ a=b \}$ the product of two conics, neither of which has rational coefficients.
Other planes, such as mathlove's choice $PB+1=2PA$ have no clear geometric meaning, but we can still rationally parametrize them and, at least in some cases, it seems we win.
As far as I can tell from skimming papers, there is an enormous literature on rational points on Kummer surfaces, but there isn't one simple answer. I had hoped to use this question as an opportunity to teach myself about Kummer surfaces (and, to some extent, I have) but it looks it's a big field, so I'll stop here.
I naively thought such a triangle would have to be equilateral, but when I tried to demonstrate it by construction I ended up producing a counterexample pretty easily.
Start by fixing points $A$ and $B$. We also fix a third point $C^\prime$ and require that $C$ lie on the ray $\vec{BC^\prime}$. Drop a perpendicular from $A$ down to $BC^\prime$ and call the point of intersection $D$. Next construct the angle bisector of $\angle ABC^\prime$; denote the intersection of the angle bisector with the line $AD$ by $G$. Finally, construct the midpoint $F$ of $AB$, and extend a line through $F$ and $G$. The intersection of $FG$ and $BC^\prime$ determines the third vertex of the triangle, $C$.
The result is that for any angle $\angle ABC^\prime$, there is a point $C$ such that the triangle $\triangle ABC$ satisfies all of our hypotheses. And of course, if $\angle ABC^\prime$ is not necessarily $60^\circ$, then $\triangle ABC$ is not necessarily equilateral.
EDIT: The last paragraph is not entirely correct. The first sentence should read: for any $\color{red}{\text{non-obtuse}}$ angle $\angle ABC^\prime$, there is a point $C$ such that the triangle $\triangle ABC$ satisfies all of our hypotheses.
Best Answer
The unit normal vector is $\vec{n}=\frac{1}{\sqrt{6}}\langle 1,-1,2\rangle$ and the vector $\overrightarrow{AB}=\langle 1,1,0\rangle$.
Using Rodrigue's rotation formula, the rotation of $\overrightarrow{AB}$ about $\vec{n}$ by $\pm 60^\circ$ is $$\overrightarrow{AC}=\frac{1}{2}\overrightarrow{AB}\pm\frac{\sqrt{3}}{2}\vec{n}\times\overrightarrow{AB}+\frac{1}{2}\vec{n}(\vec{n}\cdot \overrightarrow{AB})$$ Since we have $$\vec{n}\times\overrightarrow{AB}=\frac{1}{\sqrt{6}}\langle -2,2,2\rangle$$ $$\vec{n}\cdot\overrightarrow{AB}=0$$ Our equation for $\overrightarrow{AC}$ simplifies to \begin{align} \overrightarrow{AC} &=\frac{1}{2}\langle 1,1,0\rangle\pm\frac{\sqrt{2}}{2}\langle -1,1,1\rangle\\ &=\frac{1}{2}\langle 1,1,0\rangle -\frac{\sqrt{2}}{2}\langle -1,1,1\rangle\\ &=\boxed{\frac{1}{2}\left\langle 1+\sqrt{2},1-\sqrt{2},-\sqrt{2}\right\rangle} \end{align}