jmc

If $x^2-12x+k=0$ has two integer solutions, then $$\frac{12\pm \sqrt{144-4k}}{2}=\frac{12\pm 2\sqrt{36-k}}{2}=6\pm \sqrt{36-k}$$ must be integers. For this to be true, $36-k$ must be a perfect square. We try $36-k=1$, which means the solutions are $6\pm 1=7\text{ and }5$. These are prime numbers, so $k=\boxed{35}$ works.

Alternatively, we know that all prime numbers greater than 3 are of the form $6n-1$ or $6n+1$. So if $n=1$, we have the primes $6\pm 1=6\pm \sqrt{36-k}$. That means $\sqrt{36-k}=1$ and $k=\boxed{35}$.