Any Synthetic Solution?

I used barycentric coordinates. We get $E = (a^{2} : -b^{2} : c^{2})$ , $Q = (a^{2} : 0 : -c^{2})$ . Then $EQ$ has equation $b^{2}c^{2}x + 2a^{2}c^{2}y + a^{2}b^{2}z = 0$ . Intersecting this with the line $x = 0$ , we get $S = (0 : -b^{2} : 2c^{2})$ . We also get $T = (0 : -b^{2} : c^{2})$ . We then normalize $S$ and $T$ , and proceed to obtain $\vec{ST} = (0, \frac{b^{2}}{c^{2} - b^{2}} - \frac{b^{2}}{2c^{2} - b^{2}}, \frac{2c^{2}}{2c^{2} - b^{2}} - \frac{c^{2}}{c^{2} - b^{2}})$ . We then apply the distance formula for displacement vector $\vec{ST}$ . This gives $|\vec{ST}|^{2} = -a^{2}b^{4}c^{4}\left( \frac{1}{c^{2} -b^{2}} - \frac{1}{2c^{2} - b^{2}}\right) \left( \frac{2}{2c^{2} - b^{2}} - \frac{1}{c^{2} - b^{2}} \right) = a^{2}b^{4}c^{4}\left( \frac{1}{(c^{2} - b^{2})(2c^{2} - b^{2})} \right)^{2} \implies \\ \frac{|\vec{ST}|}{|BC|} = b^{2}c^{2}\left( \frac{1}{(c^{2} - b^{2})(2c^{2} - b^{2})} \right)$ . Substituting $b = 2c$ gives the desired value is $\\ \frac{4c^{4}}{(-3c^{2})(-2c^{2})} = \frac{2}{3}$ , so the answer is $300 \cdot \frac{2}{3}$ or $\boxed{200}$ . Let me know if there is any error.