Point $P$ is in the interior of $\triangle ABC$. The side lengths of $ABC$ are $AB = 7$, $BC = 8$, $CA = 9$. The three foots of perpendicular lines from $P$ to sides $BC$, $CA$, $AB$ are $D$, $E$, $F$ respectively. Suppose the minimal value of $\frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF}$ can be written as $\frac{a}{b}\sqrt{c}$, where $\gcd(a,b) = 1$ and $c$ is square free, calculate $abc$.
[asy]
size(120); pathpen = linewidth(0.7); pointfontpen = fontsize(10); // pointpen = black;
pair B=(0,0), C=(8,0), A=IP(CR(B,7),CR(C,9)), P = (2,1.6), D=foot(P,B,C), E=foot(P,A,C), F=foot(P,A,B);
D(A--B--C--cycle); D(P--D); D(P--E); D(P--F);
D(MP("A",A,N)); D(MP("B",B)); D(MP("C",C)); D(MP("D",D)); D(MP("E",E,NE)); D(MP("F",F,NW)); D(MP("P",P,SE));
[/asy]
1. First, we need to find the area of $\triangle ABC$ using Heron's formula. The side lengths are $AB = 7$, $BC = 8$, and $CA = 9$. The semi-perimeter $s$ is given by:
   \[
   s = \frac{AB + BC + CA}{2} = \frac{7 + 8 + 9}{2} = 12
   \]
   Using Heron's formula, the area $K$ of $\triangle ABC$ is:
   \[
   K = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{12(12-7)(12-8)(12-9)} = \sqrt{12 \cdot 5 \cdot 4 \cdot 3} = \sqrt{720} = 12\sqrt{5}
   \]

2. Let $PD = x$, $PE = y$, and $PF = z$. The areas of the smaller triangles $\triangle APB$, $\triangle BPC$, and $\triangle CPA$ can be expressed in terms of $x$, $y$, and $z$:
   \[
   [APB] = \frac{1}{2} \cdot AB \cdot PF = \frac{1}{2} \cdot 7 \cdot z = \frac{7z}{2}
   \]
   \[
   [BPC] = \frac{1}{2} \cdot BC \cdot PD = \frac{1}{2} \cdot 8 \cdot x = \frac{8x}{2}
   \]
   \[
   [CPA] = \frac{1}{2} \cdot CA \cdot PE = \frac{1}{2} \cdot 9 \cdot y = \frac{9y}{2}
   \]
   The sum of these areas is equal to the area of $\triangle ABC$:
   \[
   \frac{7z}{2} + \frac{8x}{2} + \frac{9y}{2} = 12\sqrt{5}
   \]
   Multiplying through by 2, we get:
   \[
   7z + 8x + 9y = 24\sqrt{5}
   \]

3. We need to find the minimal value of $\frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF}$, which is:
   \[
   \frac{8}{x} + \frac{9}{y} + \frac{7}{z}
   \]

4. Using the Cauchy-Schwarz inequality in the form:
   \[
   (a_1^2 + a_2^2 + a_3^2)(b_1^2 + b_2^2 + b_3^2) \geq (a_1b_1 + a_2b_2 + a_3b_3)^2
   \]
   Let $a_1 = \sqrt{8x}$, $a_2 = \sqrt{9y}$, $a_3 = \sqrt{7z}$, $b_1 = \sqrt{\frac{8}{x}}$, $b_2 = \sqrt{\frac{9}{y}}$, $b_3 = \sqrt{\frac{7}{z}}$. Then:
   \[
   (8x + 9y + 7z) \left( \frac{8}{x} + \frac{9}{y} + \frac{7}{z} \right) \geq ( \sqrt{8x} \cdot \sqrt{\frac{8}{x}} + \sqrt{9y} \cdot \sqrt{\frac{9}{y}} + \sqrt{7z} \cdot \sqrt{\frac{7}{z}} )^2
   \]
   Simplifying the right-hand side:
   \[
   (8x + 9y + 7z) \left( \frac{8}{x} + \frac{9}{y} + \frac{7}{z} \right) \geq (8 + 9 + 7)^2
   \]
   \[
   (24\sqrt{5}) \left( \frac{8}{x} + \frac{9}{y} + \frac{7}{z} \right) \geq 24^2
   \]
   \[
   24\sqrt{5} \left( \frac{8}{x} + \frac{9}{y} + \frac{7}{z} \right) \geq 576
   \]
   Dividing both sides by $24\sqrt{5}$:
   \[
   \frac{8}{x} + \frac{9}{y} + \frac{7}{z} \geq \frac{576}{24\sqrt{5}} = \frac{24}{\sqrt{5}} = \frac{24\sqrt{5}}{5}
   \]

5. The minimal value of $\frac{8}{x} + \frac{9}{y} + \frac{7}{z}$ is $\frac{24\sqrt{5}}{5}$, which can be written as $\frac{24}{5}\sqrt{5}$. Here, $a = 24$, $b = 5$, and $c = 5$. Therefore, $abc = 24 \cdot 5 \cdot 5 = 600$.

The final answer is $\boxed{600}$.