$V_O = 9-15V$ $I_L = 100mA$ $$V_O = V_{REF}(1 + \frac{R_2}{R_1}) + I _{adj}R_2$$ Assume, $V_{REF}$ = 1.25V, $R_1$ = 240Ω and $I_{adj}$ = 100µA

Let $V_O$ = 9V

$$9 = 1.25(1 + \frac{R_2}{240}) + 100µ × R_2$$ $$\boxed{R_2 = 1.45KΩ}$$ Let $V_O$ = 15V $$15 = 1.25 (1 + \frac{R_2}{240} + 100µ × R_2$$ $$\boxed{R_2 = 2.59KΩ}$$ Use potentiometer of 2.7KΩ in place of $R_2$. $$R_L = \frac{V_{omax}}{I_L}$$ $$R_L = \frac{15}{100mA}$$ $$\boxed{R_L = 150KΩ}$$

Designed circuit diagram :