A common internal tangent meets the segment between the centers at the point dividing it in the ratio of the radii. For $\mathcal{C}_1$ and $\mathcal{C}_2$ that point is $(4, 0),$ at distance $4$ from $(0,0).$ If $t_1$ makes angle $\theta$ with the $x$-axis, then $\sin\theta = \frac{1}{4},$ so $\tan\theta = \frac{1}{\sqrt{15}}$ and $t_1$ is $y = \frac{1}{\sqrt{15}}(x - 4).$ For $\mathcal{C}_2$ and $\mathcal{C}_3$ the point is $(16, 0),$ at distance $4$ from $(12, 0);$ here $\sin\theta = \frac{2}{4} = \frac{1}{2},$ so the slope is $-\frac{1}{\sqrt{3}}$ and $t_2$ is $y = -\frac{1}{\sqrt{3}}(x - 16).$

Setting the two expressions equal and multiplying by $\sqrt{15}$ gives $x - 4 = -\sqrt{5}\,(x - 16),$ so $x(1 + \sqrt{5}) = 4 + 16\sqrt{5}$ and $$\begin{aligned} x &= \frac{4 + 16\sqrt{5}}{1 + \sqrt{5}} \\ &= \frac{(4 + 16\sqrt{5})(\sqrt{5} - 1)}{4} \\ &= \frac{76 - 12\sqrt{5}}{4} \\ &= 19 - 3\sqrt{5}. \end{aligned}$$

Thus $p + q + r = 19 + 3 + 5 = 27.$