jmc

Recall that two lines are perpendicular if and only if the product of their slopes is $-1$. The first equation is already in slope-intercept form, so we can see that its slope is 2. Subtract $ax$ and divide by 6 in the second equation to get it in slope-intercept form as well: $y=-\frac{a}{6}x+1$. The negative reciprocal of 2 is $-1/2$, so setting $-a/6=-1/2$ we find that $a=\boxed{3}$ is the value for which the two lines are perpendicular.