Solution:

Answer: $(5,21,421)$

Suppose that $A$ and $B$ were rational numbers of rank $3$ less than $\frac{1}{4}$, and let $a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}$ be positive integers so that $A = \frac{1}{a_{1}} + \frac{1}{a_{2}} + \frac{1}{a_{3}}$ and $B = \frac{1}{b_{1}} + \frac{1}{b_{2}} + \frac{1}{b_{3}}$ are the expressions for $A$ and $B$ as stated in the problem. If $b_{1} < a_{1}$ then $A < \frac{1}{a_{1}-1} \leq \frac{1}{b_{1}} < B$. In other words, of all the rationals less than $\frac{1}{4}$ with rank $3$, those that have $a_{1} = 5$ are greater than those that have $a_{1} = 6,7,8, \ldots$

Therefore we can "build" $q$ greedily, adding the largest unit fraction that keeps $q$ less than $\frac{1}{4}$:

$\frac{1}{5}$ is the largest unit fraction less than $\frac{1}{4}$, hence $a_{1} = 5$;

$\frac{1}{21}$ is the largest unit fraction less than $\frac{1}{4} - \frac{1}{5}$, hence $a_{2} = 21$;

$\frac{1}{421}$ is the largest unit fraction less than $\frac{1}{4} - \frac{1}{5} - \frac{1}{21}$, hence $a_{3} = 421$.