Let $P\left(\alpha , \beta \right)$ be a point on the line $2x + 3y + 1 = 0$ such that $|PA – PB|$ is maximum where $A = (2, 0) and B = (0, 2)$ then 

Let APB $=\mathrm{\alpha }$
$\cos \mathrm{\alpha }=\frac{{\mathrm{PA}}^2+{\mathrm{PB}}^2-{\mathrm{AB}}^2}{2\mathrm{PA}\cdot \mathrm{PB}}$
$\begin{array}{l}\Rightarrow {\mathrm{PA}}^2+{\mathrm{PB}}^2-{\mathrm{AB}}^2=2\mathrm{PA}\cdot \mathrm{PBcos}\mathrm{\alpha }\le 2\mathrm{PA}\cdot \mathrm{PB}(\text{ as }\cos <1)\\ \Rightarrow (\mathrm{PA}-\mathrm{PB}{)}^2\le {\mathrm{AB}}^2\end{array}$
Thus the max value of |PA – PB| is AB
This is possible only when P lies on AB but P lies on AB
$\therefore$ P is the point of intersection of x + y = 2 and 2x + 3y + 1 = 0.

$\Rightarrow P\left(\alpha ,\beta \right)=\left(7,-5\right)$