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Let the centroid of equilateral $\triangle ABC$ be $O$ and $AO=BO=CO=r = \frac 4 {\sqrt 3}$ cm, since the side length is 4 cm. Let $O$ be the origin of the $xy$ -plane with $y$ -axis along $AO$ , the arbitrary line $l$ makes an angle $\theta$ with the $x$ -axis and the distances from vertices $A$ , $B$ and $C$ be $a$ , $b$ and $c$ respectively. Then the sum $S$ of the squares of distances from the vertices to the line is given by:

$\begin{aligned} S & = a^2 + b^2 + c^2 \\ & = r^2 \cos^2 \theta + r^2 \sin^2 (30^\circ + \theta) + r^2 \sin^2 (30^\circ - \theta) \\ & = r^2 \left(\cos^2 \theta + \sin^2 (30^\circ + \theta) + \sin^2 (30^\circ - \theta) \right) \\ & = r^2 \left(\cos^2 \theta + \sin^2 30^\circ \cos^2 \theta + 2\sin 30^\circ \cos 30^\circ \sin \theta \cos \theta + \cos^2 30^\circ \sin^2 \theta + \sin^2 30^\circ \cos^2 \theta - 2\sin 30^\circ \cos 30^\circ \sin \theta \cos \theta + \cos^2 30^\circ \sin^2 \theta \right) \\ & = r^2 \left(\cos^2 \theta + 2 \sin^2 30^\circ \cos^2 \theta + 2 \cos^2 30^\circ \sin^2 \theta \right) \\ & = r^2 \left(\cos^2 \theta + \frac 12 30^\circ \cos^2 \theta + \frac 32 \sin^2 \theta \right) \\ & = r^2 \left(\frac 32 30^\circ \cos^2 \theta + \frac 32 \sin^2 \theta \right) \\ & = \frac 32 r^2 = \frac 32 \left(\frac 4{\sqrt 3}\right)^2 = \boxed 8 \end{aligned}$