Let $f\left(x\right)=\left\{\begin{array}{ccc}ax-b& ,& x\le 1\\ 3x& ,& 1<x<2\\ b{x}^2-a& ,& x\ge 2\end{array}\right.$

If$f$ is continuous function then $(a, b)$ is equal to 

$f\left(1\right)=a-b$and $\underset{x\to 1}{lim} 3x=3$ so for $f$ to be

continuous  $a-b=3.$ Also, $f\left(2\right)=4b-a$,  so

$\underset{x\to 2-}{lim} f\left(x\right)=\underset{x\to 2-}{lim} 3x=6$ Thus $4b-a=6$

solving, we have $a = 6, b = 3.$