jmc

Let $x$ be the distance from the center $O$ of the circle to the chord of length $10$, and let $y$ be the distance from $O$ to the chord of length $14$. Let $r$ be the radius. Then, $$\begin{array}{rl}{x}^2+25& =r^2,\\ y^2+49& =r^2,\\ \mathrm{so}{x}^2+25& =y^2+49.\\ \mathrm{Therefore},x^2-y^2& =(x-y)(x+y)=24.\end{array}$$

If the chords are on the same side of the center of the circle, $x-y=6$. If they are on opposite sides, $x+y=6$. But $x-y=6$ implies that $x+y=4$, which is impossible. Hence $x+y=6$ and $x-y=4$. Solve these equations simultaneously to get $x=5$ and $y=1$. Thus, $r^2=50$, and the chord parallel to the given chords and midway between them is two units from the center. If the chord is of length $2d$, then $d^2+4=50$, $d^2=46$, and $a=(2d{)}^2=\boxed{184}$.