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

We are given the curve $$y(x-2)(x-3)=x+6$$ and we want the normal at the point where this curve meets the $$y$$-axis. On the $$y$$-axis we always have $$x=0$$, so we first substitute $$x=0$$ into the equation of the curve.

Putting $$x=0$$ gives $$y(0-2)(0-3)=0+6.$$ Evaluating the numerical factors, $$(0-2)(0-3)=(-2)(-3)=6,$$ so we obtain $$6y=6.$$ Dividing both sides by $$6$$, $$y=1.$$ Hence the curve meets the $$y$$-axis at the point $$(0,1).$$

To find the normal we need the slope of the tangent, i.e. $$\dfrac{dy}{dx}$$, at this point. The given relation is implicit, so we differentiate implicitly with respect to $$x$$.

First rewrite the left side as a product: $$(x-2)(x-3)=x^2-5x+6.$$ Thus the equation is $$y\bigl(x^2-5x+6\bigr)=x+6.$$ Let $$g(x)=x^2-5x+6.$$ The equation reads $$y\,g(x)=x+6.$$ Differentiating both sides with respect to $$x$$, remembering that $$y=y(x)$$ is a function of $$x$$, we use the product rule $$\dfrac{d}{dx}\,[y\,g(x)]=y'\,g(x)+y\,g'(x),$$ and the derivative of the right side is simply $$1.$$ Therefore $$y'\,g(x)+y\,g'(x)=1.$$ Now evaluate each factor at the point $$(0,1).$$ We have $$g(0)=(0-2)(0-3)=6,$$ and $$g'(x)=\dfrac{d}{dx}(x^2-5x+6)=2x-5,$$ so $$g'(0)=2\cdot0-5=-5.$$ Substituting $$x=0$$, $$y=1$$, $$g(0)=6$$ and $$g'(0)=-5$$ in the differentiated equation, $$y'\cdot6+1\cdot(-5)=1.$$ This simplifies step by step: $$6y'-5=1,$$ $$6y'=1+5=6,$$ $$y'=\dfrac{6}{6}=1.$$ Thus the slope of the tangent at $$(0,1)$$ is $$m_{\text{tan}}=1.$$

The slope of the normal is the negative reciprocal of the slope of the tangent. Hence $$m_{\text{norm}}=-\dfrac{1}{m_{\text{tan}}}=-\dfrac{1}{1}=-1.$$ Using the point-slope form $$y-y_1=m(x-x_1)$$ with point $$(0,1)$$ and slope $$-1$$, the equation of the normal is $$y-1=-1(x-0),$$ which rearranges to $$y=-x+1.$$

Now we test each given option to see which point satisfies $$y=-x+1$$.

Option A: $$\left(-\dfrac12,-\dfrac12\right)$$ gives $$y=-\dfrac12,$$ while $$-x+1=-(-\dfrac12)+1=\dfrac12+1=\dfrac32,$$ so it does not lie on the line.

Option B: $$\left(\dfrac12,\dfrac12\right)$$ gives $$y=\dfrac12,$$ and $$-x+1=-\dfrac12+1=\dfrac12,$$ which matches. Hence this point is on the normal.

Option C: $$\left(\dfrac12,-\dfrac13\right)$$ gives $$y=-\dfrac13,$$ whereas $$-x+1=\dfrac12,$$ so it fails.

Option D: $$\left(\dfrac12,\dfrac13\right)$$ gives $$y=\dfrac13,$$ but again $$-x+1=\dfrac12,$$ so it does not lie on the line.

Only Option B satisfies the equation of the normal. Hence, the correct answer is Option B.