How to do this surface integral? 
I first parameterise the surface as $\vec{r}(x,y)=(3x,2y,6-3x-2y)$ where $ 0 \le x \le 2 $ and  $ 0 \le y \le (6-3x)/2 $. And I am stuck here, don't know how to find the correct ${T_x} \times {T_y} $(i.e. the Jacobian). I tried to differentiate  $\vec{r}$ with respect to $x$ and $y$ respectively and compute their cross product bu that turns out to be wrong. What should I do instead?
 A: 
I first parameterise the surface as $\vec{r}(x,y)=(3x,2y,6-3x-2y)$ where $ 0 \le x \le 2 $ and  $ 0 \le y \le (6-3x)/2 $. And I am stuck here

If $3x+2y+z=6 \iff z = 6-3x-2y$, then you want to take $x$ and $y$ as parameters and then $z = 6-3x-2y$ determines $z$ as a function of $x$ and $y$. So you want
$$\vec{r}(x,y)=(\color{blue}{x},\color{blue}{y},6-3x-2y)$$
instead of
$$\vec{r}(x,y)=(\color{red}{3x},\color{red}{2y},6-3x-2y)$$

how to find the correct ${T_x} \times {T_y} $(i.e. the Jacobian). I tried to differentiate  $\vec{r}$ with respect to $x$ and $y$ respectively and compute their cross product bu that turns out to be wrong. What should I do instead?

Now you should find the correct normal vector but, since you are in fact finding a normal vector by calculating this cross product, notice that you can simply read off a normal vector from the equation of the plane:
$$\color{blue}{3}x+\color{blue}{2}y+\color{blue}{1}z=6 \implies \vec n = (\color{blue}{3},\color{blue}{2},\color{blue}{1})$$
