$=\left(\frac{9}{a^{2}}+\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+9 c^{2}\right)+4-\left(\frac{6}{a}+\frac{2 a}{b}+\frac{2 b}{c}+6 c\right)$
Now $\frac{{\frac{9}{{{a^2}}} + \frac{{{a^2}}}{{{b^2}}} + \frac{{{b^2}}}{{{c^2}}} + 9{c^2}}}{4} \ge \sqrt [4] {\frac{9}{{{a^2}}} \cdot \frac{{{a^2}}}{{{b^2}}} \cdot \frac{{{b^2}}}{{{c^2}}} \cdot 9{c^2}} \ge 3$
and $\frac{\frac{6}{a}+\frac{2 a}{b}+\frac{2 b}{c}+6 c}{4} \geq \sqrt[4]{\frac{6}{a} \cdot \frac{2 a}{b} \cdot \frac{2 b}{c} \cdot 6 c} \geq 2 \sqrt{3}$
$\therefore z \geq 12+4-8 \sqrt{3}=16-8 \sqrt{3}$
$\Rightarrow p+q+r=27$
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$\left\{(\mathrm{x}, \mathrm{y}): \frac{\mathrm{a}}{\mathrm{x}^2} \leq \mathrm{y} \leq \frac{1}{\mathrm{x}}, 1 \leq \mathrm{x} \leq 2,0<\mathrm{a}<1\right\}$ is
$\left(\log _e 2\right)-\frac{1}{7}$ then the value of $7 a-3$ is equal to: