Six equal resistances are connected between points P, Q and R as shown in the figure. Then net resistance will be maximum between :

$$R_{PQ} = r$$

$$R_{QR} = \frac{r \times r}{r + r} = \frac{r}{2}$$

$$R_{PR} = \frac{r}{3}$$

Between P and Q: $$r$$ in parallel with $$(\frac{r}{2} + \frac{r}{3} = \frac{5r}{6})$$.

$$R_{net(PQ)} = \frac{r \times \frac{5r}{6}}{r + \frac{5r}{6}} = \frac{5}{11}r \approx 0.45r$$

Between Q and R: $$\frac{r}{2}$$ in parallel with $$(r + \frac{r}{3} = \frac{4r}{3})$$.

$$R_{net(QR)} = \frac{\frac{r}{2} \times \frac{4r}{3}}{\frac{r}{2} + \frac{4r}{3}} = \frac{\frac{4r^2}{6}}{\frac{11r}{6}} = \frac{4}{11}r \approx 0.36r$$

Between P and R: $$\frac{r}{3}$$ in parallel with $$(r + \frac{r}{2} = \frac{3r}{2})$$.

$$R_{net(PR)} = \frac{\frac{r}{3} \times \frac{3r}{2}}{\frac{r}{3} + \frac{3r}{2}} = \frac{\frac{r^2}{2}}{\frac{11r}{6}} = \frac{3}{11}r \approx 0.27r$$