Let ${ }^n{C}_{r-1}=28{,}^n{C}_r=56$ and ${ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}+1}=70$. Let A(4cost, 4sint), B(2sint, –2cost) and C(3r – n, r² – n – 1) be the vertices of a triangle ABC, where t is a parameter. If $(3x-1{)}^2+(3y{)}^2=\alpha$ is the locus of the centroid of triangle ABC, then $\alpha$ equals : 

$\begin{array}{l}{ }^n{C}_{r-1}=28{,}^n{C}_r=56\\ \frac{{ }^n{C}_{r-1}}{{ }^n{C}_r}=\frac{28}{56}\\ \frac{\frac{n!}{(r-1)!(n-r+1)!}}{\frac{n!}{r!(n-r)!}}=\frac{1}{2}\\ \frac{\mathrm{r}}{(\mathrm{n}-\mathrm{r}+1)}=\frac{1}{2}\\ 3\mathrm{r}=\mathrm{n}+1............\left(i\right)\\ \frac{{ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}}{{ }^{{ }^n{\mathrm{C}}_{\mathrm{r}+1}}}=\frac{56}{70}\\ \frac{(\mathrm{r}+1)}{(\mathrm{n}-\mathrm{r})}=\frac{56}{70}\Rightarrow 9\mathrm{r}=4\mathrm{n}-5...........\left(ii\right)\\ \text{ By (i) \& (ii) }\\ (\mathrm{r}=3),(\mathrm{n}=8)\\ \mathrm{A}(4\mathrm{cost},4\mathrm{sint}), \mathrm{B}(2\mathrm{sint},-2\mathrm{cost}), \mathrm{C}\left(3\mathrm{r}-\mathrm{n},{\mathrm{r}}^2-\mathrm{n}-1\right)\\ \mathrm{A}(4\cos t,4\sin t), \mathrm{B}(2\sin t,-2\cos t), \mathrm{C}(1,0)\\ (3x-1{)}^2+(3y{)}^2=(4\cos t+2\sin t{)}^2+(4\sin t-\cos t{)}^2\\ (3x-1{)}^2+(3y{)}^2=20\end{array}$