Let $$^{n}C_{r-1}=28,^{n}C_{r}=56$$ and $$^{n}C_{r+1}=70$$. Let $$A(4\cos t,4\sin t),B(2\sin t,-2\cos t)$$ and $$C(3r-n,r^{2}-n-1)$$ 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

We are given the binomial coefficients $$\binom{n}{r-1}=28\,,\quad \binom{n}{r}=56\,,\quad \binom{n}{r+1}=70$$ and the points $$A(4\cos t,4\sin t)\,,\;B(2\sin t,-2\cos t)\,,\;C(3r-n,\;r^2-n-1)\,. $$ The locus of the centroid of triangle ABC is$$(3x-1)^2+(3y)^2=\alpha$$ and we must find $$\alpha$$.

First, from the ratio $$\frac{\binom{n}{r}}{\binom{n}{r-1}}=\frac{56}{28}=2=\frac{n-r+1}{r}$$ we get $$n-r+1=2r\quad\Rightarrow\quad n=3r-1\,. $$ Next, from $$\frac{\binom{n}{r+1}}{\binom{n}{r}}=\frac{70}{56}=\frac{5}{4}=\frac{n-r}{r+1}$$ we obtain $$4(n-r)=5(r+1)\quad\Rightarrow\quad 4n=9r+5\,. $$ Substituting $$n=3r-1$$ into $$4n=9r+5$$ gives $$4(3r-1)=9r+5\;\Rightarrow\;12r-4=9r+5\;\Rightarrow\;3r=9\;\Rightarrow\;r=3\,,\;n=8\,. $$ One checks indeed $$\binom{8}{2}=28\,,\;\binom{8}{3}=56\,,\;\binom{8}{4}=70\,. $$

With $$r=3$$ and $$n=8$$, point $$C$$ is $$C=(3r-n,\;r^2-n-1)=(9-8,\;9-8-1)=(1,0)\,. $$ The centroid $$G$$ of triangle $$ABC$$ has coordinates $$x_G=\frac{4\cos t+2\sin t+1}{3},\qquad y_G=\frac{4\sin t-2\cos t+0}{3}\,. $$ Hence $$3x_G-1=4\cos t+2\sin t,\qquad3y_G=4\sin t-2\cos t\,. $$

Therefore the locus satisfies $$(3x_G-1)^2+(3y_G)^2=(4\cos t+2\sin t)^2+(4\sin t-2\cos t)^2$$ $$=16\cos^2t+16\sin t\cos t+4\sin^2t+16\sin^2t-16\sin t\cos t+4\cos^2t$$ $$=16(\cos^2t+\sin^2t)+4(\sin^2t+\cos^2t)=16+4=20\,. $$

Hence $$\alpha=20$$, which corresponds to Option D.