jmc

Let $f(x)=x^2+2ax+3a.$ Then we want the graph of $y=f(x)$ to intersect the "strip" $-2\le y\le 2$ in exactly one point. Because the graph of $y=f(x)$ is a parabola opening upwards, this is possible if and only if the minimum value of $f(x)$ is $2.$

To find the minimum value of $f(x),$ complete the square: $$f(x)=(x^2+2ax+a^2)+(3a-a^2)=(x+a{)}^2+(3a-a^2).$$It follows that the minimum value of $f(x)$ is $3a-a^2,$ so we have $$3a-a^2=2,$$which has solutions $a=\boxed{1,2}.$