The inradius of \(\triangle ABC\) is \(100\sqrt{3}\) and the circumradius is \(200\sqrt{3}\). Consider the line perpendicular to plane \(ABC\) through the circumcenter of \(\triangle ABC\). Note that \(P, Q, O\) must lie on that line to be equidistant from each of the triangle's vertices. Also, note that since \(P, Q, O\) are collinear, and \(OP = OQ\), we must have \(O\) is the midpoint of \(PQ\). Now, Let \(K\) be the circumcenter of \(\triangle ABC\), and \(L\) be the foot of the altitude from \(A\) to \(BC\). We must have \(\tan(\angle KLP + \angle QLK) = \tan(120°)\). Setting \(KP = x\) and \(KQ = y\), assuming WLOG \(x > y\), we must have \[\tan(120°) = -\sqrt{3} = \frac{\dfrac{x+y}{100\sqrt{3}}}{\dfrac{30000 - xy}{30000}}.\] Thus \(100(x+y) = xy - 30000\). Also, \(\left(\dfrac{x+y}{2}\right)^2 = \left(\dfrac{x-y}{2}\right)^2 + 120000\) by the Pythagorean theorem, so \(xy = 120000\), and substituting, \(90000 = 100(x+y)\), or \(x + y = 900\). The desired answer is \(\dfrac{x+y}{2}\).