Tricky Triplet Trepidation

We can rewrite the equation as

$(3a + 8b + 36c)^2 = 13^2(a^2 + 4b^2 + 9c^2).$

But, by the Cauchy-Schwarz Inequality,

$\begin{aligned} (3a + 8b + 36c)^2 &= (3 \cdot a + 4 \cdot 2b + 12 \cdot 3c)^2 \\ &\leq (3^2 + 4^2 + 12^2)(a^2 + (2b)^2 + (3c)^2) \\ &= 13^2(a^2 + 4b^2 + 9c^2). \end{aligned}$

Thus, we must have the equality case for the inequality, which in this case is $\dfrac{a}{3} = \dfrac{2b}{4} = \dfrac{3c}{12}.$ This gives $\dfrac{a}{c} = \dfrac{3}{4}$ and $\dfrac{b}{c} = \dfrac{1}{2},$ so $\dfrac{a + b}{c} = \dfrac{5}{4},$ and $p + q = \boxed{9}.$

@Steven Yuan A nice problem!!!! Well, we can also solve this using vectors and a small substitution.......!!!!