If  $f\left(x\right)=(x^2-3x+2)(x^2-6x+p)$ and  $g\left(x\right)=(x^2-4x+3)(x^2-8x+q)$, then the value of $a$ and  $b$ , if $(x-1)(x-2)$ is  $\text{H.C.F}$ of  $f\left(x\right)$ and  $g\left(x\right)$ is

$f\left(x\right)=(x^2-3x+2)(x^2-6x+p)$

$=(x-1)(x-2)(x^2-6x+p)$

$g\left(x\right)=(x^2-4x+3)(x^2-8x+q)$

$g\left(x\right)=(x-1)(x-3)(x^2-8x+q)$

Given  $\text{H.C.F}$ is  $(x-1)(x-2)$

$\therefore$ The possible value of  $\text{'}p\text{'}$ is any real value  $p\in R$

If  $(x^2-8x+q)$ is divisible by  $(x-2)$

$\begin{array}{l}\text{then\hspace{0.17em}\hspace{0.17em}}4-16+q=0\\ q=12\end{array}$