(ii) 2n C3 : n C3 = 11:1

The given ratio is $^{2n}C_3 : ^{n}C_3 = 11:1$, which can be written as:

$$\frac{^{2n}C_3}{^{n}C_3} = \frac{11}{1}$$

Substitute the combination formula:

$$\frac{\frac{(2n)!}{3!(2n-3)!}}{\frac{n!}{3!(n-3)!}} = 11$$

$$\frac{(2n)!}{3!(2n-3)!} \times \frac{3!(n-3)!}{n!} = 11$$

Cancel out $3!$:

$$\frac{(2n)!}{(2n-3)!} \times \frac{(n-3)!}{n!} = 11$$

Expand the factorials: $(2n)! = 2n(2n-1)(2n-2)(2n-3)!$ and $n! = n(n-1)(n-2)(n-3)!$.

$$\frac{2n(2n-1)(2n-2)(2n-3)!}{(2n-3)!} \times \frac{(n-3)!}{n(n-1)(n-2)(n-3)!} = 11$$

Cancel out $(2n-3)!$ and $(n-3)!$:

$$\frac{2n(2n-1)(2n-2)}{n(n-1)(n-2)} = 11$$

Factor out 2 from $(2n-2)$:

$$\frac{2n(2n-1)2(n-1)}{n(n-1)(n-2)} = 11$$

Cancel out $n$ and $(n-1)$ (assuming $n \ge 3$ for the combinations to be defined):

$$\frac{4(2n-1)}{n-2} = 11$$

$$4(2n-1) = 11(n-2)$$ $$8n - 4 = 11n - 22$$

Rearrange the terms to solve for $n$:

$$22 - 4 = 11n - 8n$$ $$18 = 3n$$ $$n = \frac{18}{3}$$ $$n = 6$$

The value of $n$ is $\boxed{\text{6}}$.