Better method to solve a geometric problem.
From $\triangle BPS \sim \triangle CPA$ one obtains $$BS = \frac{AC\cdot BP}{CP} = \frac{45z}{2y}$$ and from $\triangle AQT \sim \triangle CQB$ one gets $$AT = \frac{BC\cdot AQ}{CQ} = \frac{7y}{2z}.$$ Hence $$BS \cdot AT = \frac{315}{4}.$$ We then get by Ptolemy theorem $$BS \cdot AT = AS \cdot BT + AB \cdot ST,$$ or equivalently $$\frac{315}{4}=5\cdot 2+11\cdot ST.$$ It follows that $$ST = \frac{\frac{315}{4}-10}{11} = \frac{25}{4}.$$