Solution:

Since $PA = PB = PC$ and $QA = QB = QC,$ both $P$ and $Q$ lie on the line through the center $H$ of $\triangle ABC$ perpendicular to its plane, on opposite sides. Any point equidistant from $A,$ $B,$ $C$ also lies on that line, so $O$ is on it, and $OP = OQ = d$ makes $O$ the midpoint of $\overline{PQ},$ with $PQ = 2d.$ Let $D$ be the midpoint of $\overline{AB}$ and $a = 600;$ then $DH = \frac{a\sqrt{3}}{6}$ and $CH = \frac{a\sqrt{3}}{3}.$ Since $\overline{PD} \perp \overline{AB}$ and $\overline{QD} \perp \overline{AB},$ the dihedral angle is $\angle PDQ = 120^\circ;$ write $x = \angle PDH$ and $y = \angle QDH,$ so $x + y = 120^\circ.$

Right triangles $PDH$ and $QDH$ give $PH = DH \tan x$ and $QH = DH \tan y,$ so $$2d = PQ = PH + QH = \frac{a\sqrt{3}}{6}(\tan x + \tan y).$$ Since $OC = OP = OQ = d,$ point $C$ lies on the circle with diameter $\overline{PQ},$ so $\angle PCQ = 90^\circ,$ and $H$ is the foot of the altitude from $C$ to the hypotenuse $\overline{PQ}.$ Thus $CH^2 = PH \cdot QH,$ which gives $\tan x \tan y = \frac{CH^2}{DH^2} = 4.$

By the tangent addition formula, $-\sqrt{3} = \tan 120^\circ = \frac{\tan x + \tan y}{1 - \tan x \tan y} = \frac{\tan x + \tan y}{-3},$ so $\tan x + \tan y = 3\sqrt{3}.$ Then $2d = \frac{a\sqrt{3}}{6} \cdot 3\sqrt{3} = \frac{3a}{2},$ so $d = \frac{3a}{4} = 450.$