Let the circum centre of a triangle with vertices $\mathrm{A}\left(\mathrm{a},3\right), \mathrm{B}\left(\mathrm{b},5\right) \mathrm{and} \mathrm{C}\left(\mathrm{a},\mathrm{b}\right), \mathrm{ab}>0$ be $\mathrm{P}\left(1,1\right)$. If the line AP intersects the line BC at the point $Q\left(k_1,k_2\right)$, then $k_1+k_2$  is equal to :

$\begin{array}{r}\mathrm{Mid} \mathrm{point} \mathrm{of} \mathrm{AC}= \mathrm{D}\left(\frac{\mathrm{a}+\mathrm{a}}{2},\frac{\mathrm{b}+3}{2}\right)\Rightarrow \mathrm{D}\left(\mathrm{a},\frac{\mathrm{b}+3}{2}\right)\\ \mathrm{Slope} \mathrm{of} \mathrm{AC}=\frac{3-\mathrm{b}}{0}\Rightarrow \mathrm{Slope} \mathrm{of} \mathrm{PD}={\mathrm{m}}_{\mathrm{PD}}=0\\ \Rightarrow \frac{\mathrm{b}+3}{2}-1=0\therefore \boxed{\mathrm{b}=-1}\\ \mathrm{Mid} \mathrm{point} \mathrm{of} \mathrm{BC}=\mathrm{E}\left(\frac{\mathrm{b}+\mathrm{a}}{2},\frac{5+\mathrm{b}}{2}\right)=\left(\frac{a-1}{2},2\right)\\ \begin{array}{r}{\mathrm{m}}_{\mathrm{CB}}\cdot {\mathrm{m}}_{\mathrm{EP}}=-1\Rightarrow \left(\frac{5-\mathrm{b}}{\mathrm{b}-\mathrm{a}}\right)\left(\frac{2-1}{\frac{\mathrm{a}-1}{2}-1}\right)=-1\\ \begin{array}{r}\Rightarrow {\mathrm{a}}^2-2\mathrm{a}-15=0\Rightarrow (\mathrm{a}-5)(\mathrm{a}+3)=0\\ \sin ce ab>0,\boxed{ a=-3}\end{array}\end{array}\end{array}$

$$\begin{gathered} \mathrm{A}\left(-3,3\right),\mathrm{P}\left(1,1\right)\Rightarrow \mathrm{Equation} \mathrm{of} \mathrm{AP} \mathrm{is} \boxed{\mathrm{x}+2\mathrm{y}=3} \\ \mathrm{and} \mathrm{B}\left(-1,5\right),\mathrm{C}\left(-3,-1\right)\Rightarrow \mathrm{Equation} \mathrm{ofBC} \mathrm{is} \boxed{\mathrm{y}=3\mathrm{x}+8} \\ \mathrm{Point} \mathrm{of} \mathrm{intersection} \mathrm{of} \mathrm{the} \mathrm{lines} \mathrm{AP} \mathrm{and} \mathrm{BC} \mathrm{is} \left(\frac{-13}{7},\frac{17}{7}\right) \\ \therefore k_1+k_2=\frac{-13+17}{7}=\frac{4}{7} \end{gathered}$$