Polynomial of $11^{th}$ degree

HINT:

Let $(x+1)f(x)=1+A\prod_{r=0}^{11}(x-r)$ where $A$ is an arbitrary constant

Somewhat less vague...

$(x+1)f(x)-1$ is a polynomial of degree 12 with roots at every integer in $[0,11]$, so could be $$(x+1)f(x)-1 = A \prod_{c=0}^{11} x-c$$ for some/any (nonzero) constant $A$.

When $x=-1$, we have $(0)f(-1)-1 = A (-1)^{12} 12!$, or $-1 = A \, 12!$ and discover only $A = \frac{-1}{12!}$ is consistent with the givens. (Why $-1$? Because it is the only choice we haven't already used (we have used the integers 0, ... 11) that makes some expression containing $x$s zero.)

Hence, $13 f(12) - 1 = \frac{-1}{12!} 12!$ and $f(12) = \frac{-1+1}{13} = 0$.