The normal to the curve  $y\left(x-2\right) \left(x-3\right)=x+6$ at the point where the curve intersects the $y$-axis passes through the point:

To find the point where curve intersects the y-axis 

Put, $x=0$, we get

 $6y=6\Rightarrow y=1$

Hence, the point is (0, 1).

On differentiating

$\Rightarrow \left(\frac{dy}{dx}\right) \left(x-2\right) \left(x-3\right)+y\left(2x-5\right)=1$

Put, $x=0, y=1$, we get

$\Rightarrow \frac{dy}{dx}\left(+6\right)+1\left(-5\right)=1$

$\Rightarrow \frac{dy}{dx}=1$

$\Rightarrow$ Slope of normal $=-1$

Equation of normal at $\left(0, 1\right)$

$\Rightarrow y-1=-1\left(x-0\right)$

$\Rightarrow x+y-1=0$

Point $\left(\frac{1}{2},\frac{1}{2}\right)$ satisfy the normal.