Let $x,y,z\in\mathbb{R^+}$ , such that

Note that $\frac{9}{4+5+6}=\frac35$ and $\frac{12}{4+5+6} = \frac45$ , while $3^2+4^2=5^2$ . Thus, if we define $x \; = \; 4 \times \tfrac35 \hspace{1cm} y \; = \; 5 \times \tfrac35 \hspace{1cm} z \; = \; 6 \times \tfrac35$ then $\sqrt{16-x^2} \; = \; 4 \times \tfrac45 \hspace{1cm} \sqrt{25-y^2} \; = \; 5 \times \tfrac45 \hspace{1cm} \sqrt{36-z^2} \; = \; 6 \times \tfrac45$ and hence $x+y+z \; = \; (4+5+6)\tfrac35 \; =\; 9 \hspace{2cm} \sqrt{16-x^2} + \sqrt{25-y^2} + \sqrt{36-z^2} \; = \; (4+5+6)\tfrac45 \; =\; 12$ which means that we want $xyz \; = \; 4 \times 5 \times 6 \times \big(\tfrac35\big)^3 \; = \; \tfrac{648}{25} \; = \; \boxed{25.92}$