jmc

Because the coefficients of the polynomial are rational, the radical conjugate of $3-\sqrt{7}$, which is $3+\sqrt{7}$, must also be a root of the polynomial. By Vieta's formulas, the sum of the roots of this polynomial is $0$; since $(3-\sqrt{7})+(3+\sqrt{7})=6,$ the third, integer root must be $0-6=\boxed{-6}.$