The angles in a particular triangle are in arithmetic progression, and the side lengths are $6,7,x$. The sum of the possible values of $x$ equals $a+\sqrt b+\sqrt c$, where $a,b,c$ are positive integers. What is $a+b+c$?

1. Since the angles are in AP, the middle angle is $60^\circ$. If $60^\circ$ is opposite $x$: $x^2=36+49-42=43$.
2. If $60^\circ$ is opposite the larger given side ($7$): $49=36+x^2-6x\Rightarrow x^2-6x-13=0\Rightarrow x=3+\sqrt{22}$ (the case with $60^\circ$ opposite $6$ gives no real solution).
3. Sum $=\sqrt{43}+3+\sqrt{22}$, so $a+b+c=3+22+43=\boxed{68}$.