Let ${ }^n C_{r-1}=28,{ }^n C_r=56$ and ${ }^n C_{r+1}=70$. Let $A(4 \operatorname{cost}, 4 \sin t), B(2 \sin t,-2 \cos t)$ and $C\left(3 r-n, r^2-n-1\right)$ be the vertices of a triangle $A B C$, where $t$ is a parameter. If $(3 x-1)^2+(3 y)^2$ $=\alpha$, is the locus of the centroid of triangle ABC , then $\alpha$ equals

<p>$$\begin{aligned} & { }^n C_{r-1}=28,{ }^n C_r=56 \\ & \frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}}=\frac{28}{56} \\ & \frac{\frac{n!}{(r-1)!(n-r+1)!}}{\frac{n!}{r!(n-r)!}}=\frac{1}{2} \\ & \frac{\mathrm{r}}{(\mathrm{n}-\mathrm{r}+1)}=\frac{1}{2} \\ & 3 \mathrm{r}=\mathrm{n}+1\quad\text{..... (i)} \end{aligned}$$</p> <p>$$\begin{aligned} & \frac{{ }^n C_r}{{ }^n C_{r+1}}=\frac{56}{70} \\ & \frac{(r+1)}{(n-r)}=\frac{56}{70} \Rightarrow 9 r=4 n-5\quad\text{.... (ii)} \end{aligned}$$</p> <p>$$\begin{aligned} & \text { By (i) & (ii) } \\ & \begin{array}{l} (r=3),(n=8) \\ A(4 \cos t, 4 \sin t) \quad B(2 \sin t,-2 \cos t) C\left(3 r-n, r^2-n-1\right) \\ A(4 \cos t, 4 \sin t) \quad B(2 \sin t,-2 \operatorname{cost}) C(1,0) \\ (3 x-1)^2+(3 y)^2=(4 \operatorname{cost}+2 \sin t)^2+(4 \sin t-\operatorname{cost})^2 \\ (3 x-1)^2+(3 y)^2=20 \quad \therefore \text { option }(1) \end{array} \end{aligned}$$</p>