$\underset{x\to 2}{\lim }\frac{3x^2-ax+5b}{x-2}=17$

Since, limit exists, numerator must be zero at $x=2$

$\Rightarrow 12-2a+5b=0$

$\Rightarrow 5b+12=2a$

Using L'Hospital's rule, we get

$\underset{x\to 2}{\lim }\frac{6x-a}{1}=17$

$\Rightarrow 6x=17+a$

$\Rightarrow a=-5$

So, $5b+12=2\left(-5\right)$

$\Rightarrow b=\frac{-22}{5}$

So, $ab=\left(-5\right)\left(\frac{-22}{5}\right)$

Hence, $ab=22$