jmc

Since this must hold for any quadratic, let's look at the case where $p(x)=x^2.$ Then the given equation becomes $$n^2=r(n-1{)}^2+s(n-2{)}^2+t(n-3{)}^2.$$This expands as $$n^2=(r+s+t)n^2+(-2r-4s-6t)n+r+4s+9t.$$Matching the coefficients on both sides, we get the system $$\begin{array}{rl}r+s+t& =1,\\ -2r-4s-6t& =0,\\ r+4s+9t& =0.\end{array}$$Solving this linear system, we find $r=3,$ $s=-3,$ and $t=1.$

We verify the claim: Let $p(x)=a{x}^2+bx+c.$ Then $$\begin{array}{rl}& 3p(n-1)-3p(n-2)+p(n-3)\\ & =3(a(n-1{)}^2+b(n-1)+c)-3(a(n-2{)}^2+b(n-2)+c)+a(n-3{)}^2+b(n-3)+c\\ & =a(3(n-1{)}^2-3(n-2{)}^2+(n-3{)}^2)+b(3(n-1)-3(n-2)+(n-3))+c(3-3+1)\\ & =a{n}^2+bn+c\\ & =p(n).\end{array}$$Thus, the claim is true, and $(r,s,t)=\boxed{(3,-3,1)}.$