The coordinates of the points A and B are respectively (–3,2) and (2, 3). P and Q are points on the line joining A and B such that AP = PQ = QB. A square PQRS is constructed on PQ as one side, the coordinates of R can be

$P\left(-\frac{4}{3},\frac{7}{3}\right)\text{ and }Q\left(\frac{1}{3},\frac{8}{3}\right)\text{, then }PQ=\frac{\sqrt{26}}{3}$

Let the coordinates of R be (x, y).

Then, $QR=PQ\Rightarrow {\left(x-\frac{1}{3}\right)}^2+{\left(y-\frac{8}{3}\right)}^2=\frac{26}{9}$ …….(i)

and QR is perpendicular to PQ 

$\Rightarrow \frac{\mathrm{y}-\frac{8}{3}}{\mathrm{x}-\frac{1}{3}}\times \frac{\frac{8}{3}-\frac{7}{3}}{\frac{1}{3}+\frac{4}{3}}=-1$…..(ii)

From Eqs.(i) and (ii), we get

$26{\left(x-\frac{1}{3}\right)}^2=\frac{26}{9},x-\frac{1}{3}=\pm \frac{1}{3}$

$\Rightarrow x=0\text{ or }x=\frac{2}{3}$From Eq.(ii),

when $x=0,y=\frac{13}{3}$ and when $x=\frac{2}{3},y=1$. So, 

the required coordinates of R are $\left(0,\frac{13}{3}\right)\text{ or }\left(\frac{2}{3},1\right)$