smc

Angle $C$ is split into three $30^{\circ }$ angles. The shorter angle trisector will be the one closer $BC$. Let it intersect $AB$ at point $P$. Let the perpendicular from point $P$ intersect $BC$ at point $R$ and have length $x$. Thus $△PRC$ is a $30^{\circ }-60^{\circ }-90^{\circ }$ triangle and $RC$ has length $x\sqrt{3}$. Because $△PBR$ is similar to $△ABC$, $RB$ has length $\frac{3}{4}x$. $$RC+RB=BC=x\sqrt{3}+\frac{3}{4}x=3$$ The problem asks for the length of $PC$, or $2x$. Solving for $x$ and multiplying by two gives $\boxed{\frac{32\sqrt{3}-24}{13}}$.