My Very Own Twist of a Problem!

$a-b-c=\log _{ 10 }{ 12 } -\log _{ 10 }{ 6 } -\log _{ 10 }{ \sqrt { 2 } } =\log _{ 10 }{ 2 } -\frac { \log _{ 10 }{ 2 } }{ 2 } =\log _{ 10 }{ \sqrt { 2 } } \\ { 10 }^{ \log _{ 10 }{ \sqrt { 2 } } }=\sqrt { 2 } =1.44\\ 0<{ 9 }^{ \log _{ 10 }{ \sqrt { 2 } } }<{ 10 }^{ \log _{ 10 }{ \sqrt { 2 } } }\\ The\quad only\quad option\quad that\quad satisfies\quad that\quad is\quad \boxed{A:1.39}$

The given values can be rewritten as

$10^a=12 \\ 10^b=6 \\ 10^c=\sqrt{2} \Rightarrow 10^{2c}=2$

As we know $6\times 2=12$

$\therefore 10^b\times 10^{2c}=10^a$

$\large10^{b+2c}=10^a$

$\therefore b+2c=a \Rightarrow a-b=2c$

we have to find $\large 9^{a-b-c} \Rightarrow 9^{2c-c} \Rightarrow 9^{c}$

$\large 9^{\log{2}^{\frac{1}{2}}} \Rightarrow 3^{\log{2}} \Rightarrow 1.39$

Hence our answer is $\boxed{1.39}$

${a = \log_{10}12, b = \log_{10}6, c = \log_{10} \sqrt{2}}$ ${{a - b - c} = {\log_{10}12} - {\log_{10}6} - {\log_{10} \sqrt{2}} = {\log_{10}\sqrt{2}}}$ ${{9}^{\log_{10}\sqrt{2}} = {3}^{\log_{10}2} \approx \boxed {1.39}}$

With these options, it's easy to find the right one without manipulation. a is a little over one, b somewhat under one, and c can't be much at all. 1.39 is the only answer remotely between 9^0 and 9^1.