If we let $a = |BC|$, $b = |CA|$ and $c = |AB|$ and use that $\cos(120^\circ) = -1/2$, the Cosine Rule gives $49 = a^2 = b^2 + c^2 + bc$. Moreover, the area of triangle $ABC$ is equal to $\frac{7}{2} \cdot \frac{15\sqrt{3}}{14} = \frac{15\sqrt{3}}{4}$ and this is equal to $\frac{bc\sin(120^\circ)}{2} = \frac{bc\sqrt{3}}{4}$, hence $bc = 15$. We obtain $b^2 + c^2 = 49 - bc = 34$ and so
$$(b+c)^2 = b^2 + 2bc + c^2 = 34 + 30 = 64 \\ (b-c)^2 = b^2 - 2bc + c^2 = 34 - 30 = 4.$$
Therefore, $b + c = 8$ and $b - c = \pm 2$. This leads to $(b, c) = (3, 5)$ or $(b, c) = (5, 3)$ and we see that the lengths of the other two sides of triangle $ABC$ are 3 and 5.