In $\triangle \mathrm{ABC}$, the angles $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and their opposite sides $\mathrm{a}, \mathrm{b}, \mathrm{c}$ respectively. Given that the area of $\triangle \mathrm{ABC}$ is $3\sqrt{15}$, $b - c = 2$, $\cos A = -\frac{1}{4}$, then the value of $a$ is \underline{\hspace{2cm}}.