The problem reads:
Compute $\int_S \, \vec{F}(x, y, z) \cdot \vec{n} \ dS$, where $\vec{F}(x, y, z) = \left\langle x\ln(xz), 5z, \frac{1}{y^2+1} \right\rangle$, $S$ is the region of the plane $12x-9y+3z=10$ over the rectangular region in the $xy$-plane $D=\{(x, y) \; | \; 2 \le x \le 3 \; \text{and} \; 5 \le y \le 10\}$, and $\vec{n}$ points upwards.
The surface $S$ is defined by $z=f(x,y)=\frac{10}{3}-4x+3y$.
The differential is $dS=\sqrt{\left(\frac{\partial{f}}{\partial{x}}\right)^2+\left(\frac{\partial{f}}{\partial{y}}\right)^2+1} \, dA = \sqrt{26} \, dA$.
The normal vector is $\vec{n} = \frac{1}{\sqrt{26}} \left\langle 4, -3, 1 \right\rangle$.
Since $z = \frac{10}{3}-4x+3y$,
$$\vec{F}\left[x, y, \left( \frac{10}{3} - 4x + 3y \right) \right] = \left\langle x\ln\left( \frac{10}{3}x - 4x^2 + 3xy \right), \left( \frac{50}{3} - 20x + 15y \right), \frac{1}{y^2+1} \right\rangle$$
$$\begin{align} \int_S \, \vec{F}(x, y, z) \cdot \vec{n} \ dS & = \int_2^3 \int_5^{10} \left( 4x \ln\left[\frac{10}{3}x - 4x^2 + 3xy\right] - 50 + 60x - 45y + \frac{1}{y^2+1} \right) dy \, dx \end{align}$$
It seems that this double integral becomes tedious to solve. Are there any mistakes I made in the steps towards setting up the problem?
It doesn't look like you made any mistakes. Maybe whoever setup the problem got confused and thought that $\vec n=(0,0,1)$? That would have made the integral trivial.