Let $\rm AS_a\cap BC\equiv K,AG\cap BC \equiv M,Q\equiv KG\cap AD$ and $\rm AD\perp BC,D\in BC$ Since $\rm GG_a=GS_a $ we have that $\rm QD=QA$.Also $\rm \dfrac{KG}{KQ}=\dfrac{GG_a}{DQ}=2\dfrac{GG_a}{AD}=2\dfrac{MG}{MA}=\dfrac{2}{3}$ so $\rm G$ is the centoid of $\rm ADK$.Therefore $\rm DM=MK$. Hence ,the lines $\rm AS_a,BS_b,CS_c$ are concurrent at the isotomic conjugate point of orthocenter of triangle $\rm ABC$.

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Can this problem be solved using the fact MSb = MG = MSc? So, I added the point La which is symmetrical point of G respect to Ga. Using this new point, I proved MSb = MG = MSc but I don't know how to finish the task.