jmc

Expanding, we get $$\begin{array}{rl}(x+i)((x+1)+i)((x+2)+i)& =(x^2+x+xi+(x+1)i+i^2)((x+2)+i)\\ & =(x^2+x-1+(2x+1)i)((x+2)+i)\\ & =(x^2+x-1)(x+2)+(x^2+x-1)i+(2x+1)(x+2)i+(2x+1)i^2\\ & =(x^3+3x^2-x-3)+(3x^2+6x+1)i\end{array}$$We want this complex number to be pure imaginary, so the real part $x^3+3x^2-x-3$ must be 0. This factors as $$(x+3)(x+1)(x-1)=0,$$so the solutions are $\boxed{-3,-1,1}.$