1. Equation of line $AB$

Points are $A(-1,1)$ and $B(2,3)$.

Slope of $AB$ is $m=\frac{3-1}{2-(-1)}=\frac{2}{3}.$

So equation of line $AB$ is $y-1=\frac{2}{3}(x+1).$

Simplifying, $3y-3=2x+2$ $2x-3y+5=0.$

2. Using area condition

For a variable point $P(x,y)$, area of triangle $PAB$ is $\frac{1}{2}\times AB \times (\text{perpendicular distance of }P\text{ from line }AB).$

Given area $=10$, so perpendicular distance from $P$ to line $AB$ is constant.

First find length $AB$: $AB=\sqrt{(2+1)^2+(3-1)^2}=\sqrt{3^2+2^2}=\sqrt{13}.$

Hence, $10=\frac{1}{2}\cdot \sqrt{13}\cdot d$ $d=\frac{20}{\sqrt{13}}.$

3. Locus of point at constant distance from line $AB$

Distance of $(x,y)$ from line $2x-3y+5=0$ is $\frac{|2x-3y+5|}{\sqrt{2^2+(-3)^2}}=\frac{|2x-3y+5|}{\sqrt{13}}.$

Set this equal to $\dfrac{20}{\sqrt{13}}$: $\frac{|2x-3y+5|}{\sqrt{13}}=\frac{20}{\sqrt{13}}$ $|2x-3y+5|=20.$

So the two possible lines are $2x-3y+5=20 \quad \text{or} \quad 2x-3y+5=-20.$

That is, $2x-3y-15=0$ or $2x-3y+25=0.$

4. Choose the line above $AB$

We need the line corresponding to points above line $AB$.

Take a point clearly above $AB$, for example $(0,3)$.

For line $AB: 2x-3y+5=0$, $2(0)-3(3)+5=-4<0.$

So points above $AB$ satisfy $2x-3y+5<0.$

Thus we choose $2x-3y+5=-20$ $2x-3y=-25.$

To match the form $ax+by=15$, multiply by $-\frac{3}{5}$: $-\frac{6}{5}x+\frac{9}{5}y=15.$

Hence, $a=-\frac{6}{5}, \qquad b=\frac{9}{5}.$

5. Compute $5a+2b$

$5a+2b=5\left(-\frac{6}{5}\right)+2\left(\frac{9}{5}\right)=-6+\frac{18}{5}=-\frac{30}{5}+\frac{18}{5}=-\frac{12}{5}.$

6. Option check

The correct option is $\boxed{A\;\left(-\frac{12}{5}\right)}.$

7. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

So they agree.