1. Use the fact about the greatest binomial coefficient

For a fixed $n$, the binomial coefficient ${}^nC_r$ is greatest when $r$ is nearest to $\dfrac n2$.

• If $n$ is odd, the two middle coefficients are equal and greatest.
• If $n$ is even, the single middle coefficient is greatest.

So:

• For ${}^{19}C_p$, the greatest value is at $p=9$ or $10$: $a = {}^{19}C_9 = {}^{19}C_{10}$

• For ${}^{20}C_q$, the greatest value is at $q=10$: $b = {}^{20}C_{10}$

• For ${}^{21}C_r$, the greatest value is at $r=10$ or $11$: $c = {}^{21}C_{10} = {}^{21}C_{11}$

2. Relate $b$ and $a$

Using Pascal's identity: ${}^{20}C_{10} = {}^{19}C_{10} + {}^{19}C_9$

But since ${}^{19}C_{10} = {}^{19}C_9 = a,$ we get $b = a+a = 2a.$

Hence, $\frac{b}{22} = \frac{2a}{22} = \frac{a}{11}.$

3. Relate $c$ and $b$

Again by Pascal's identity: ${}^{21}C_{10} = {}^{20}C_{10} + {}^{20}C_9$

Now for $n=20$, symmetry gives ${}^{20}C_9 = {}^{20}C_{11}.$ Also, ${}^{20}C_{10} = \frac{20!}{10!10!}, \qquad {}^{20}C_9 = \frac{20!}{9!11!}.$

So, $\frac{{}^{20}C_{10}}{{}^{20}C_9} = \frac{11}{10}$ which means ${}^{20}C_9 = \frac{10}{11}b.$

Therefore, $c = {}^{21}C_{10} = b + \frac{10}{11}b = \frac{21}{11}b.$

Thus, $\frac{c}{42} = \frac{1}{42}\cdot \frac{21}{11}b = \frac{b}{22}.$

So, $\frac{a}{11} = \frac{b}{22} = \frac{c}{42}.$

4. Check options

The correct relation is $\frac{a}{11} = \frac{b}{22} = \frac{c}{42}.$

So the correct option is D.