The correct order of relative basic strength will be
$\ddot{\mathrm{NH}}_3>\ddot{\mathrm{NH}}_2-\ddot{\mathrm{NH}}_2>\ddot{\mathrm{NH}}_2-\mathrm{OH}$
In $\mathrm{NH}_3$, the lone pair is completely available for donation, hence is most basic whereas $\mathrm{NH}_2-\mathrm{NH}_2$ and $\mathrm{NH}_2-\mathrm{OH}$ can be considered as derivatives of $\mathrm{NH}_3$. Here, $\mathrm{H}$ is replaced by $-\mathrm{NH}_2$ forming $\mathrm{NH}_2-\mathrm{NH}_2$ and $-\mathrm{OH}$ forming $\mathrm{NH}_2-\mathrm{OH}$ being highly electron withdrawing due to $-\mathrm{O}$ will decrease the election density most on $-\mathrm{N}$.
$\therefore$ Least basic elect;on withdrawing $-I$-effect of
$-\mathrm{NH}_2$ is less as compared to - $\mathrm{OH}$.
Hence correct order of basic strength is
$\mathrm{NH}_3>\mathrm{NH}_2-\mathrm{NH}_2>\mathrm{NH}_2-\mathrm{OH}$