Let P be a point on the parabola, $$y^2 = 12x$$ and N be the foot of the perpendicular drawn from P, on the axis of the parabola. A line is now drawn through the mid-point M of PN, parallel to its axis which meets the parabola at Q. If the $$y$$-intercept of the line NQ is $$\frac{4}{3}$$, then:

The equation $$y^{2}=12x$$ can be compared with the standard parabola $$y^{2}=4ax$$. Equating the coefficients we obtain $$4a=12$$, so $$a=3$$.

For a parabola of the form $$y^{2}=4ax$$ the parametric coordinates of any point are given by the well-known formula $$P(at^{2},\,2at)$$. Substituting $$a=3$$ we write the point $$P$$ on the given parabola as $$P\;(3t^{2},\,6t).$$

The axis of this parabola is the $$x$$-axis, i.e. the line $$y=0$$. The foot of the perpendicular drawn from $$P$$ to the axis therefore has the same $$x$$-coordinate as $$P$$ and a $$y$$-coordinate of zero. Hence $$N\;(3t^{2},\,0).$$

The length of the segment $$PN$$ is simply the absolute difference of their $$y$$-coordinates because the segment is vertical: $$PN=\bigl|6t-0\bigr|=6|t|.$$

Now we need the mid-point of $$PN$$. Using the mid-point formula $$M\left(\frac{x_{P}+x_{N}}{2},\,\frac{y_{P}+y_{N}}{2}\right),$$ we substitute $$P(3t^{2},6t)$$ and $$N(3t^{2},0)$$ to get $$M\left(\frac{3t^{2}+3t^{2}}{2},\,\frac{6t+0}{2}\right) =\bigl(3t^{2},\,3t\bigr).$$

Through $$M$$ we draw a line parallel to the axis of the parabola. Because the axis is the $$x$$-axis, a parallel line is a horizontal line. Thus the required line is simply $$y=3t.$$

This horizontal line meets the parabola again at a point $$Q$$. To find the coordinates of $$Q$$ we substitute $$y=3t$$ in the parabola’s equation $$y^{2}=12x$$:

$$\bigl(3t\bigr)^{2}=12x \;\Longrightarrow\;9t^{2}=12x \;\Longrightarrow\;x=\frac{9}{12}t^{2}=\frac{3}{4}t^{2}.$$

Hence $$Q\left(\frac{3}{4}t^{2},\,3t\right).$$

Observe that $$M$$ and $$Q$$ have the same $$y$$-coordinate. The distance $$MQ$$ is therefore purely horizontal: $$MQ=\bigl|x_{M}-x_{Q}\bigr| =\Bigl|3t^{2}-\frac{3}{4}t^{2}\Bigr| =\left(3-\frac{3}{4}\right)t^{2} =\frac{9}{4}\,t^{2}.$$

Next we use the information about the line $$NQ$$. First we find its slope. Using the two points $$N(3t^{2},\,0),\qquad Q\!\left(\frac{3}{4}t^{2},\,3t\right),$$ the slope is

$$m=\frac{3t-0}{\dfrac{3}{4}t^{2}-3t^{2}} =\frac{3t}{\left(\dfrac{3}{4}-3\right)t^{2}} =\frac{3t}{\left(\dfrac{3}{4}-\dfrac{12}{4}\right)t^{2}} =\frac{3t}{-\dfrac{9}{4}t^{2}} =-\frac{4}{3}\,\frac{1}{t}.$$

Using point-slope form, the equation of $$NQ$$ is $$y-0=m\bigl(x-3t^{2}\bigr) \;\Longrightarrow\;y=-\frac{4}{3t}\,\bigl(x-3t^{2}\bigr).$$

The $$y$$-intercept is obtained by putting $$x=0$$:

$$y=-\frac{4}{3t}\,(0-3t^{2}) =-\frac{4}{3t}\,(-3t^{2}) =\frac{12t^{2}}{3t} =4t.$$

We are told that this intercept equals $$\dfrac{4}{3}$$, so

$$4t=\frac{4}{3}\;\Longrightarrow\;t=\frac{1}{3}.$$

With $$t=\dfrac{1}{3}$$ we now evaluate $$MQ$$:

$$MQ=\frac{9}{4}\,t^{2} =\frac{9}{4}\left(\frac{1}{3}\right)^{2} =\frac{9}{4}\cdot\frac{1}{9} =\frac{1}{4}.$$

Thus $$MQ=\dfrac{1}{4}$$.

Looking at the given options, this corresponds to Option C.

Hence, the correct answer is Option C.