jmc

Adding $4$ to both sides, we have $$\left(1+\frac{3}{x-3}\right)+\left(1+\frac{5}{x-5}\right)+\left(1+\frac{17}{x-17}\right)+\left(1+\frac{19}{x-19}\right)=x^2-11x$$or $$\frac{x}{x-3}+\frac{x}{x-5}+\frac{x}{x-17}+\frac{x}{x-19}=x^2-11x.$$Either $x=0$, or $$\frac{1}{x-3}+\frac{1}{x-5}+\frac{1}{x-17}+\frac{1}{x-19}=x-11.$$To induce some symmetry, we calculate that the average of the numbers $x-3,x-5,x-17,x-19$ is $x-11$. Then, letting $t=x-11$, we have $$\frac{1}{t+8}+\frac{1}{t+6}+\frac{1}{t-6}+\frac{1}{t-8}=t,$$or, combining the first and last terms and the second and third terms, $$\frac{2t}{t^2-64}+\frac{2t}{t^2-36}=t.$$Either $t=0$, or we can divide by $t$ and cross-multiply, giving $$2(t^2-36)+2(t^2-64)=(t^2-36)(t^2-64)⟹0=t^4-104t^2+2504.$$Completing the square, we get $(t^2-52{)}^2=200$, so $t^2=52\pm \sqrt{200}$, and $t=\pm \sqrt{52\pm \sqrt{200}}$. Undoing the substitution $t=x-11$, we have $$x=11\pm \sqrt{52\pm \sqrt{200}}.$$Therefore, the largest root is $x=11+\sqrt{52+\sqrt{200}}$ (which is larger than both $x=0$ and $t=0⟹x=11$), and the answer is $11+52+200=\boxed{263}$.