Problem 147

(I) From $\cos 2C - 3\cos C = 1$, we get $2\cos^2 C - 1 - 3\cos C = 1$, which simplifies to $2\cos^2 C - 3\cos C - 2 = 0$. This factors to $(2\cos C + 1)(\cos C - 2) = 0$.

However, since $\cos C \leq 1$, it follows that $\cos C - 2 < 0$, and thus $\cos C = -\frac{1}{2}$.

Since $C$ is an internal angle of a triangle, $0 < C < \pi$. Therefore, $C = \frac{2\pi}{3}$.

(II) From $S_{\triangle ABC} = \frac{1}{2}ab\sin C = \frac{3\sqrt{3}}{2}$ and $C = \frac{2\pi}{3}$, we get $ab = 6$.

Using the cosine rule, we have $c^2 = a^2 + b^2 - 2ab\cos C = (a + b)^2 - 2ab(1 + \cos C)$. Substituting the given values, we get $7 = (a + b)^2 - 12(1 - \frac{1}{2})$, which simplifies to $a + b = \sqrt{13}$.

According to the sine rule, $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$. Therefore, $\sin A + \sin B = \frac{a\sin C}{c} + \frac{b\sin C}{c} = \frac{(a + b)\sin C}{c} = \frac{\sqrt{13} \times \frac{\sqrt{3}}{2}}{\sqrt{7}} = \frac{\sqrt{39}}{2\sqrt{7}}$.

Hence, $\boxed{\sqrt{7}(\sin A + \sin B) = \frac{\sqrt{39}}{2}}$.