No Cubic Discriminant Involved

Let the roots be $a,b,c$ where $a,b$ are integers. $a+b+c=0$ $\Rightarrow c=-(a+b)$ $ab+bc+ca=-27$ $\Rightarrow ab-(a+b)^2=-27$ $\Rightarrow b^2-ab+a^2-27=0$ $\Rightarrow \triangle = 108-3a^2$ Since $b$ is an integer, the discriminant is a perfect square. It has a factor of $3$ already, so $a$ must also be a multiple of $3$ .

The discriminant is also non-negative, so $a^2 \leq 36$ $\Rightarrow a = -6,3,0,3,6$ Trying all values of $a$ , all the solutions are $(a,b,c)=(-6,3,3),(-3,-3,6),(-3,6,-3),(3,-6,3),(3,3,-6),(6,-3,-3)$ $\Rightarrow k = \boxed{-54 \mbox{ or } 54}$