Much the same as Imo said. The tetrahedron has four faces and to find the flux directly you will need to integrate over all four faces.

Three of the faces are in the coordinate planes. Of course, the plane x+ 2y+ 3z= 6 crosses the three axes at (6, 0, 0), (0, 3, 0) and (0, 0, 2) so the face in the xy plane (z= 0) is the triangle with vertices (0, 0, 0), (6, 0, 0), and (0, 3, 0). \(\displaystyle d\vec{S}= <0, 0, 1>dxdy\) so you need to integrate <x, y, z>.<0, 0, 1>dxdy= zdxdy= 0 (because z= 0 in this plane) over that triangle which just gives 0. Similarly for the other two sides in the coordinate planes.

That is why you really only need to integrate on the x+ 2y+ 3z= 6 face. That can be written x= 6- 2y- 3z or, as a vector formula, \(\displaystyle \vec{r}(y,z)= (6- 2y- 3z)\vec{i}+ y\vec{j}+ z\vec{k}\). The derivative with resepect to y is \(\displaystyle -2\vec{i}+ \vec{j}\) and with respect to z, \(\displaystyle -3\vec{i}+ \vec{k}\), so the "\(\displaystyle \partial r/\partial u\times \partial r/\partial v\)" of Imo is

\(\displaystyle \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k}\\ -2 & 1 & 0 \\ -3 & 0 & 1\end{array}\right|= \vec{i}+ 2\vec{j}+ 3\vec{k}\)

Integrate \(\displaystyle (x\vec{i}+ y\vec{j}+ z\vec{k})\cdot(\vec{i}+ 2\vec{j}+ 3\vec{k})dydz\) over the triangle, in the yz-plane, with vertices at (0, 0, 0), (0, 3, 0), and (0, 0, 2).