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

We have triangle with vertices $$A(a, 3)$$, $$B(b, 5)$$, $$C(a, b)$$ with $$ab > 0$$, and circumcentre $$P(1, 1)$$. Since $$P$$ is the circumcentre, we have $$PA^2 = PB^2 = PC^2$$.

We compute each squared distance. We have $$PA^2 = (a-1)^2 + (3-1)^2 = (a-1)^2 + 4$$, and $$PB^2 = (b-1)^2 + (5-1)^2 = (b-1)^2 + 16$$, and $$PC^2 = (a-1)^2 + (b-1)^2$$.

From $$PA^2 = PC^2$$, we get $$(a-1)^2 + 4 = (a-1)^2 + (b-1)^2$$, which gives $$(b-1)^2 = 4$$, so $$b = 3$$ or $$b = -1$$.

From $$PB^2 = PC^2$$, we get $$(b-1)^2 + 16 = (a-1)^2 + (b-1)^2$$, which gives $$(a-1)^2 = 16$$, so $$a = 5$$ or $$a = -3$$.

Now we apply the constraint $$ab > 0$$ (both must have the same sign) and also check that the triangle is non-degenerate.

Case 1: $$a = 5, b = 3$$. Then $$A = (5, 3)$$ and $$C = (5, 3)$$, so $$A = C$$. This is degenerate, so we discard it.

Case 2: $$a = 5, b = -1$$. Then $$ab = -5 < 0$$. This violates $$ab > 0$$, so we discard it.

Case 3: $$a = -3, b = 3$$. Then $$ab = -9 < 0$$. This violates $$ab > 0$$, so we discard it.

Case 4: $$a = -3, b = -1$$. Then $$ab = 3 > 0$$ ✓, and the vertices are $$A(-3, 3)$$, $$B(-1, 5)$$, $$C(-3, -1)$$, which form a valid triangle.

Now we find the line $$AP$$. We have $$A(-3, 3)$$ and $$P(1, 1)$$. The slope is $$\frac{1 - 3}{1 - (-3)} = \frac{-2}{4} = -\frac{1}{2}$$. The equation of line $$AP$$ is $$y - 3 = -\frac{1}{2}(x + 3)$$, which simplifies to $$y = -\frac{x}{2} + \frac{3}{2}$$.

Next we find the line $$BC$$. We have $$B(-1, 5)$$ and $$C(-3, -1)$$. The slope is $$\frac{-1 - 5}{-3 - (-1)} = \frac{-6}{-2} = 3$$. The equation of line $$BC$$ is $$y - 5 = 3(x + 1)$$, which simplifies to $$y = 3x + 8$$.

Setting the two equations equal to find $$Q$$: $$-\frac{x}{2} + \frac{3}{2} = 3x + 8$$. Multiplying through by 2: $$-x + 3 = 6x + 16$$, so $$-7x = 13$$, giving $$x = -\frac{13}{7}$$.

Substituting back: $$y = 3\left(-\frac{13}{7}\right) + 8 = -\frac{39}{7} + \frac{56}{7} = \frac{17}{7}$$.

Hence $$Q = \left(-\frac{13}{7},\, \frac{17}{7}\right)$$, and $$k_1 + k_2 = -\frac{13}{7} + \frac{17}{7} = \frac{4}{7}$$.

Hence, the correct answer is Option B.