jmc

Let the three roots be $r-d$, $r$, and $r+d$, for some complex numbers $r$ and $d$. Then Vieta's formulas give $$(r-d)+r+(r+d)=6\text{and}(r-d)r+(r-d)(r+d)+r(r+d)=21.$$Simplifying these equations, we have $$3r=6\text{and}3r^2-d^2=21.$$From $3r=6$, we deduce $r=2$. Substituting this into our second equation gives $12-d^2=21$, so $d^2=-9$ and $d=\pm 3i$. Therefore, the roots of the cubic are $2-3i$, $2$, and $2+3i$, so $$a=-2(2-3i)(2+3i)=-2\left(2^2-(3i{)}^2\right)=-2(4+9)=\boxed{-26}.$$