Let the circumcentre of a triangle with vertices A (a, 3), B(b, 5)...

On plotting the graph we get,

Now by diagram ${m}_{AC} ⟶ \infty$ and ${m}_{PD} = 0$ as $PD$ is perpendicular bisector, so $D\left(\dfrac{a + a}{2}, \dfrac{b + 3}{2}\right)$

$≡ D\left(a, \dfrac{b + 3}{2}\right)$

Now since ${m}_{PD} = 0$, so $\dfrac{b + 3}{2} - 1 = 0$

$⇒ b + 3 - 2 = 0$$⇒ b = - 1$

Now similarly finding the coordinates of $E\left(\dfrac{b + a}{2}, \dfrac{5 + b}{2}\right) = \left(\dfrac{a - 1}{2}, 2\right)\left\{\text{as}b = - 1\right\}$

Now we know that product of perpendicular slopes is ${m}_{CB} \cdot {m}_{EP} = - 1$

$⇒ \left(\dfrac{5 - b}{b - a}\right) = \left(\dfrac{2 - 1}{\dfrac{a - 1}{2} - 1}\right) = - 1$

$⇒ \left(\dfrac{6}{- 1 - a}\right) = \left(\dfrac{2}{a - 3}\right) = - 1$

$⇒ 12 = \left(1 + a\right)\left(a - 3\right)$

$⇒ 12 = {a}^{2} - 3a + a - 3$

$⇒ {a}^{2} - 2a - 15 = 0$

$⇒ \left(a - 5\right)\left(a + 3\right) = 0$

So, $a = 5$ or $a = - 3$

Now given $ab > 0$

So, $a\left(- 1\right) > 0$$⇒ a < 0$

So, $a = - 3$

Now equation of line $AP$ by two point form will be, {where $A\left(- 3, 3\right) \text{ and } P\left(1, 1\right)$}

$y - 1 = \left(\dfrac{3 - 1}{- 3 - 1}\right)\left(x - 1\right)$

$⇒ - 2y + 2 = x - 1$

$⇒ x + 2y = 3$ $⋯ \left(1\right)$

Now equation of line $BC$ {where $B\left(- 1, 5\right)$ and $C\left(- 3, - 1\right)$} will be,

\left(y - 5\right) = \dfrac{6}{2}$\left(x + 1\right)$

$⇒ y - 5 = 3x + 3$

$⇒ y = 3x + 8 ⋯ \left(2\right)$

Now solving \left(1\right)$and \left(2\right)$ we get,

$x + 2\left(3x + 8\right) = 3$

$⇒ 7x + 16 = 3$$⇒ 7x = - 13$

$⇒ x = - \dfrac{13}{7}$

And $y = 3\left(- \dfrac{13}{7}\right) + 8$$= \dfrac{- 39 + 56}{7}$

So, $y = \dfrac{17}{7}$

So the value of \left({k}_{1}, {k}_{2}\right)$will be$x + y = \dfrac{- 13 + 17}{7} = \dfrac{4}{7}$\left\{\text{where}{k}_{1}\text{is}x \text{ and } {k}_{2}\text{is}y\right\}$