jmc

Solving $2y+x+3=0$ for $y$ gives $y=\frac{-1}{2}x-\frac{3}{2},$ so the slope of this line is $-\frac{1}{2}.$

Solving $3y+ax+2=0$ for $y$ gives $y=\frac{-a}{3}x-\frac{2}{3},$ so the slope of this line is $-\frac{a}{3}.$

In order for these lines to be perpendicular, we must have $$\left(-\frac{1}{2}\right)\left(-\frac{a}{3}\right)=-1.$$Solving for $a$ gives $a=\boxed{-6}.$