Three Lines, Three Circles v2

same analysis, you post it well

It is clear that the circles described in the problem are the excircles of $ABC.$ Let $r_{a}, r_{b},$ and $r_{c}$ be the radii of these excircles. Also, let $r$ be the inradius of $ABC.$ It can be proven that

$[ABC] = \sqrt{rr_{a}r_{b}r_{c}},$

where $[ABC]$ denotes the area of $ABC.$ We already know the exradii lengths. To get the inradius, we use the fact that

$\frac{1}{r_{a}} + \frac{1}{r_{b}} + \frac{1}{r_{c}} = \frac{1}{r}.$

From this, we get

$\begin{aligned} r &= \frac{1}{\frac{1}{r_{a}} + \frac{1}{r_{b}} + \frac{1}{r_{c}}} \\ &= \frac{1}{\frac{1}{3} + \frac{1}{4} + \frac{1}{5}} \\ &= \frac{1}{\frac{47}{60}} \\ &= \frac{60}{47} \end{aligned}$

and

$\begin{aligned} [ABC] &= \sqrt{rr_{a}r_{b}r_{c}} \\ &= \sqrt{\left (\frac{60}{47} \right )(3)(4)(5)} \\ &= \frac{60}{\sqrt{47}} \\ &= \frac{60\sqrt{47}}{47}. \end{aligned}$

Thus, $p = 60, q = 47,$ and $r = 47,$ and $p + q + r = \boxed{154}.$