If ${\left(1-\sqrt{2}-\sqrt{3}\right)}^{-1}=a+b\sqrt{2}+c\sqrt{6}$  then

On rationalizing denominator

$=\frac{1+\left(\sqrt{2}+\sqrt{3}\right)}{1^2-{\left(\sqrt{2}+\sqrt{3}\right)}^2}$

$=\frac{1+\sqrt{2}+\sqrt{3}}{-\left(4+2\sqrt{6}\right)}\times \frac{4-2\sqrt{6}}{4-2\sqrt{6}}$

$=\frac{\left(1+\sqrt{2}+\sqrt{3}\right)\left(4-2\sqrt{6}\right)}{-\left[4^2-{\left(2\sqrt{6}\right)}^2\right]}$

$=\frac{4-2\sqrt{2}-2\sqrt{6}}{8}$

$=a+b\sqrt{2}+c\sqrt{6}$

$\therefore a=\frac{1}{2},b=\frac{-1}{4},c=\frac{-1}{4}$