Is $x=y=z=0$ ?

$\begin{cases} (x-2)(y-2) = 4 & ... (1) \\ (y-3)(z-3) = 9 & ... (2) \\ (z-4)(x-4) = 16 & ...(3) \end{cases}$

$\begin{aligned} (1): \ \quad y-2 & = \frac{4}{x-2} \\ \implies y & = \frac{4}{x-2} + 2 \\ (3): \quad \implies z & = \frac{16}{x-4} + 4 \end{aligned}$

$\begin{aligned} (2): \quad \left(\frac{4}{x-2} - 1\right) \left(\frac{16}{x-4} + 1 \right) & = 9 \\ \left(\frac{4 - x + 2}{x-2}\right) \left(\frac{16+x-4}{x-4} \right) & = 9 \\ \left(\frac{6 - x}{x-2}\right) \left(\frac{12+x}{x-4} \right) & = 9 \\ (6-x)(12+x) & = 9(x-2)(x-4) \\ -x^2 -6x+72 & = 9x^2-54x+72 \\ 10x^2 - 48x & = 0 \\ x(5x-24) & = 0 \\ \implies x & = \frac{24}{5} \quad \quad \small \color{#3D99F6}{x \ne 0 \text{, since }x \text{ is positive rational}} \end{aligned}$

$\implies y = \dfrac{4}{\frac{24}{5}-2} + 2 = \dfrac{20}{14} + 2 = \dfrac{24}{7}$

$\implies z = \dfrac{16}{\frac{24}{5}-4} + 4 = \dfrac{80}{4} + 4 = 24$

Therefore, $5x+7y-z = 5\left(\dfrac{24}{5}\right) + 7\left(\dfrac{24}{7}\right) - 24 = \boxed{24}$