If  ${\cos }^{-1}x+{\cos }^{-1}2x+{\cos }^{-1}3x=\pi$and  $a{x}^3+b{x}^2+cx=0$ satisfies the equation  then the value of $\frac{b-a-c}{100}$ is 

$$\begin{gathered} {\cos }^{-1}x+{\cos }^{-1}2x+{\cos }^{-1}3x=\pi \\ \Rightarrow {\cos }^{-1}2x+{\cos }^{-1}3x=\pi -{\cos }^{-1}x={\cos }^{-1}\left(-x\right) \\ \Rightarrow 6x^2-\sqrt{1-4x^2}\sqrt{1-9x^2}=-x \\ \Rightarrow 6x^2+x=\sqrt{1-4x^2}\sqrt{1-9x^2} \\ \Rightarrow 36x^4+12x^3+x^2=1-13x^2+36x^4 \\ \Rightarrow 12x^3+14x^2-1=0 \end{gathered}$$

$$\begin{gathered} \Rightarrow a=12,b=14,c=-1 \\ \therefore \frac{b-a-c}{100}=\frac{3}{100}=0.03 \end{gathered}$$