It forms a circle of radius \$\sqrt{3/2} \approx 1.224745\$ centered on origin, with a diagonal axis (towards \$(-1,-1,-1)\$).
This is the biggest circle you can fit in a cube with vertices at \$(\pm 1, \pm 1, \pm 1)\$ according to e.g
this essay by Maris Ozols. (Note that Ozol's hypercube had unit length sides, whereas here we have sides of length 2 and \$n = 3\$; and that \$2 \sqrt{3/8} = \sqrt{3/2}\$.)
How can I tell? I asked Maxima:
load("vect") $
"Helper functions:" $
radians(deg) := deg * %pi/180 $
cross(a,b) := express(a ~ b) $
"P(deg) is point 'deg' degrees along the curve." $
P(deg) := [ sin(radians(deg)), sin(radians(deg + 120)), sin(radians(deg + 240)) ] $
"Pick three points along the curve; any three distinct points will do. Better pick points where sine has a known value." $
p0 : P(0) $
p1 : P(120) $
p2 : P(240) $
"We know the origin is the center by definition. Find normal." $
n : cross(p1 - p0, p2 - p0) $
ratsimp(n);
"Check the distances of the three points from origin." $
ratsimp([sqrt(p0 . p0), sqrt(p1 . p1), sqrt(p2 . p2)]);
float(%);
If you choose e.g.
p0 : P(90) $ p1 : P(180) $ p2 : P(270) $ you get the exact same answer (except for the magnitude of the normal vector, which is irrelevant here, as it depends on the angle
p1 p0 p2 ).