Let $$A(1, 1)$$, $$B(-4, 3)$$, $$C(-2, -5)$$ be vertices of a triangle $$ABC$$, $$P$$ be a point on side $$BC$$, and $$\Delta_1$$ and $$\Delta_2$$ be the areas of triangle $$APB$$ and $$ABC$$ respectively. If $$\Delta_1 : \Delta_2 = 4 : 7$$, then the area enclosed by the lines $$AP$$, $$AC$$ and the $$x$$-axis is

The vertex of the parabola is $$V = (5, 4)$$ and the directrix is $$3x + y - 29 = 0$$.

The axis of the parabola is perpendicular to the directrix with direction $$(3, 1)$$.

Distance from $$V$$ to the directrix: $$\dfrac{|3(5) + 4 - 29|}{\sqrt{9+1}} = \dfrac{|-10|}{\sqrt{10}} = \sqrt{10}$$.

Since $$3(5)+4-29 = -10 < 0$$, the vertex is on the side where $$3x+y < 29$$. The focus lies on the same side, at distance $$\sqrt{10}$$ from the vertex along the direction $$-(3,1)/\sqrt{10}$$.

Focus: $$S = (5,4) - \sqrt{10} \cdot \dfrac{(3,1)}{\sqrt{10}} = (5-3, 4-1) = (2, 3)$$.

By the definition of a parabola, for any point $$P(x,y)$$: distance to focus = distance to directrix.

$$(x-2)^2 + (y-3)^2 = \dfrac{(3x+y-29)^2}{10}$$

Multiplying both sides by 10:

$$10(x^2 - 4x + 4 + y^2 - 6y + 9) = (3x+y-29)^2$$

$$10x^2 - 40x + 40 + 10y^2 - 60y + 90 = 9x^2 + 6xy - 174x + y^2 - 58y + 841$$

Rearranging: $$x^2 + 9y^2 - 6xy + 134x - 2y - 711 = 0$$

So $$a = 9$$, $$b = -6$$, $$c = 134$$, $$d = -2$$, $$k = -711$$.

$$a + b + c + d + k = 9 - 6 + 134 - 2 - 711 = -576$$.

The answer is Option D: $$-576$$.