In $\Delta ABC,$ side $AC$ and the perpendicular bisector of $BC$ meet at $D$, where $BD$ bisects $\angle ABC$.

Let $E=(0,0)$, $B=(-\alpha,0)$, $C=(\alpha,0)$, $D=(0,\beta)$. Then we have $$\alpha^2+\beta^2=7^2=49,\quad\tan \angle ACB=\frac{\beta}{\alpha}\in(0,\sqrt3).$$ Let $t=\tan \angle ACB$, then $$y_{AB}=\frac{2t}{1-t^2}(x+\alpha),\quad y_{AC}=-tx+\beta.$$ And we have $$A=\left(-\alpha\frac{1+t^2}{3-t^2},\alpha\frac{4t}{3-t^2}\right),\quad |AB|=2\alpha\frac{1+t^2}{3-t^2}.$$ Thus, the area \begin{align} S_{\triangle ABD} &= \frac12|AB|\cdot|DE|\\ &=\alpha\beta\frac{1+t^2}{3-t^2}\\ &=\frac{49t}{3-t^2}\in(0,\infty). \end{align} If you have some other conditions such that $\frac{t}{3-t^2}=\sqrt5$, then $a=49$. If not, $a$ can be any positive real number.

Note:- As others mentioned, this question was unsolvable before because there can be infinitely many such triangles. I also realized that is true because I missed a small detail that $AD = 9$. Sorry for the inconvenience. I also figured out how to solve this now.

I had figured out that :- $$\frac{k}{m} = \frac{7}{2n} = \frac{m}{(7+k)}$$ With the new information this becomes :- $$\frac{9}{m} = \frac{7}{2n} = \frac{m}{16}$$

From here, I get $m = 12$ , so I have the lengths of all the sides of $\Delta ABC$.
Now from Heron's Formula I have :- $$[\Delta ABD] = \sqrt{s(s-a)(s-b)(s-c)}$$ $$\rightarrow \sqrt{14 * 2 * 5 * 7}$$ $$\rightarrow \sqrt{980} = 14\sqrt{5}$$ So we have $a = 14$ as our solution.