Let $$A(-1, 1)$$ and $$B(2, 3)$$ be two points and $$P$$ be a variable point above the line $$AB$$ such that the area of $$\triangle PAB$$ is 10. If the locus of $$P$$ is $$ax + by = 15$$, then $$5a + 2b$$ is :

Given points $$A(-1, 1)$$ and $$B(2, 3)$$, and P is a variable point above line AB such that area of $$\triangle PAB = 10$$.

First, determine the equation of line AB. Its slope is $$\frac{3-1}{2-(-1)} = \frac{2}{3}$$, giving $$y - 1 = \frac{2}{3}(x + 1)$$, which simplifies to $$3y - 3 = 2x + 2$$ and hence $$2x - 3y + 5 = 0$$.

The length of AB is $$\sqrt{(2-(-1))^2 + (3-1)^2} = \sqrt{9 + 4} = \sqrt{13}$$.

Using the area formula for a triangle, $$\frac{1}{2} \times AB \times d = 10$$, where $$d$$ is the perpendicular distance from P to line AB, we have $$10 = \frac{1}{2} \times \sqrt{13} \times d$$, so $$d = \frac{20}{\sqrt{13}}$$.

Let $$P = (x,y)$$. The distance from P to the line $$2x - 3y + 5 = 0$$ is $$\frac{|2x - 3y + 5|}{\sqrt{4+9}} = \frac{|2x - 3y + 5|}{\sqrt{13}}$$. Setting this equal to $$\frac{20}{\sqrt{13}}$$ yields $$|2x - 3y + 5| = 20$$.

Since P is above AB, the expression $$2x - 3y + 5$$ must be negative—for instance, at (0, 10), it equals -25—so $$2x - 3y + 5 = -20$$. This gives $$2x - 3y = -25$$, or equivalently $$2x - 3y + 25 = 0$$, which can also be written as $$-2x + 3y = 25$$.

To express this in the form $$ax + by = 15$$, multiply $$2x - 3y = -25$$ by $$-\frac{3}{5}$$ to obtain $$-\frac{6}{5}x + \frac{9}{5}y = 15$$, giving $$a = -\frac{6}{5}$$ and $$b = \frac{9}{5}$$.

Finally, $$5a + 2b = 5\Bigl(-\frac{6}{5}\Bigr) + 2\Bigl(\frac{9}{5}\Bigr) = -6 + \frac{18}{5} = \frac{-30 + 18}{5} = -\frac{12}{5}$$, so the correct answer is Option (4): $$-\frac{12}{5}$$.