Given that $x^2+x-6$  is a factor of $2x^4+x^3-a{x}^2+bx+a+b-1,$  then the value of $a-3b$  is ___.

$x^2+x-6=(x+3)(x-2)$ 

Let $f\left(x\right)=2x^4+x^3-a{x}^2+bx+a+b-1.$

 Therefore, $f\left(-3\right)=2{\left(-3\right)}^4+{\left(-3\right)}^3-a{\left(-3\right)}^2-3b+a+b-1=0$

$\Rightarrow 134-8a-2b=0$

$\Rightarrow 4a+b=67$       -------$\left(1\right)$

and $f\left(2\right)=2{\left(2\right)}^4+2^3-a{\left(2\right)}^2+2b+a+b-1=0$

$\Rightarrow 39-3a+3b=0$

$\Rightarrow a-b=13$           ------- $\left(2\right)$

From (1) and (2) we get $a=16,b=3.$