The area of the region $$\left\{(x, y): y^2 \leq 4x, x < 4, \frac{xy(x-1)(x-2)}{(x-3)(x-4)} > 0, x \neq 3\right\}$$ is

The inequality $$y^{2} \le 4x$$ represents the right-opening parabola $$y = \pm 2\sqrt{x}$$.
For every admissible $$x$$ the vertical length of the full strip would be $$2\sqrt{x} - (-2\sqrt{x}) = 4\sqrt{x}$$, but the second condition will reduce this length by half. We analyse that condition next.

Let $$g(x) = \dfrac{x(x-1)(x-2)}{(x-3)(x-4)}$$. The given sign restriction can be rewritten as

$$y \, g(x) \gt 0 \qquad -(1)$$

Equation $$(1)$$ says that $$y$$ and $$g(x)$$ must have the same sign. Hence, for a fixed $$x$$ only one half of the total $$y$$-interval survives: if $$g(x) \gt 0$$ we keep $$0 \lt y \le 2\sqrt{x}$$; if $$g(x) \lt 0$$ we keep $$-2\sqrt{x} \le y \lt 0$$.
In either case the surviving vertical length is $$2\sqrt{x}$$.

Because $$y^{2} \le 4x$$ requires $$x \ge 0$$, and because $$x \lt 4$$ is already imposed, we only have to study the sign of $$g(x)$$ on the intervals obtained by the zeros and poles of $$g(x)$$:

$$0,\,1,\,2,\,3,\,4$$ split the axis (with $$x=3$$ excluded). A sign table gives

In all four intervals the admissible vertical length is $$2\sqrt{x}$$, so the required area $$A$$ equals

$$ A = 2 \int_{0}^{1} \!\sqrt{x}\,dx + 2 \int_{1}^{2} \!\sqrt{x}\,dx + 2 \int_{2}^{3} \!\sqrt{x}\,dx + 2 \int_{3}^{4} \!\sqrt{x}\,dx \qquad -(2) $$

Using the antiderivative $$\displaystyle\int \sqrt{x}\,dx = \frac{2}{3}x^{3/2}$$ in $$(2)$$:

$$ \begin{aligned} A &= \frac{4}{3}\Bigl[x^{3/2}\Bigr]_{0}^{1} + \frac{4}{3}\Bigl[x^{3/2}\Bigr]_{1}^{2} + \frac{4}{3}\Bigl[x^{3/2}\Bigr]_{2}^{3} + \frac{4}{3}\Bigl[x^{3/2}\Bigr]_{3}^{4} \\[2mm] &= \frac{4}{3}\Bigl(1 - 0\Bigr) + \frac{4}{3}\Bigl(2^{3/2} - 1\Bigr) + \frac{4}{3}\Bigl(3^{3/2} - 2^{3/2}\Bigr) + \frac{4}{3}\Bigl(4^{3/2} - 3^{3/2}\Bigr) \\[2mm] &= \frac{4}{3}\Bigl[1 + (2\sqrt{2}-1) + (3\sqrt{3}-2\sqrt{2}) + (8-3\sqrt{3})\Bigr] \\[2mm] &= \frac{4}{3}\,(8) \\[2mm] &= \frac{32}{3}. \end{aligned} $$

Therefore, the area of the given region is $$\dfrac{32}{3}$$, which is Option D.