1. Find $n$ and $r$ from the binomial coefficients

Given: $^nC_{r-1}=28,\, ^nC_r=56,\, ^nC_{r+1}=70$

Use the ratio: $\frac{^nC_r}{^nC_{r-1}}=\frac{n-r+1}{r}$ So, $\frac{56}{28}=2=\frac{n-r+1}{r}$ $n-r+1=2r$ $n+1=3r \quad \Rightarrow \quad n=3r-1$

Now use: $\frac{^nC_{r+1}}{^nC_r}=\frac{n-r}{r+1}$ So, $\frac{70}{56}=\frac{5}{4}=\frac{n-r}{r+1}$ $4(n-r)=5(r+1)$ $4n-4r=5r+5$ $4n=9r+5$

Substitute $n=3r-1$: $4(3r-1)=9r+5$ $12r-4=9r+5$ $3r=9$ $r=3$ Then, $n=3(3)-1=8$

So, $r=3, \quad n=8$

2. Find coordinates of point $C$

Given: $C(3r-n,\, r^2-n-1)$ Substitute $r=3, n=8$: $C(3\cdot 3-8,\, 3^2-8-1)=(1,0)$

3. Coordinates of centroid of triangle $ABC$

Given: $A(4\cos t,\, 4\sin t), \quad B(2\sin t,\, -2\cos t), \quad C(1,0)$

Let centroid be $G(x,y)$. Then $x=\frac{4\cos t+2\sin t+1}{3}$ $y=\frac{4\sin t-2\cos t}{3}$

So, $3x-1=4\cos t+2\sin t$ $3y=4\sin t-2\cos t$

4. Find the locus

We need: $(3x-1)^2+(3y)^2$ Substitute: $=(4\cos t+2\sin t)^2+(4\sin t-2\cos t)^2$

Expand both terms: $=16\cos^2 t+16\sin t\cos t+4\sin^2 t+16\sin^2 t-16\sin t\cos t+4\cos^2 t$

Cross terms cancel: $=(16\cos^2 t+4\cos^2 t)+(4\sin^2 t+16\sin^2 t)$ $=20\cos^2 t+20\sin^2 t$ $=20(\cos^2 t+\sin^2 t)=20$

Hence the locus is $(3x-1)^2+(3y)^2=20$ Therefore, $\alpha=20$

5. Compare with stored answer

Derived answer: Option C, $20$.

Stored correct answer: C.

They match.