<s>[INST] Are the boundaries of the integral correct? I want to calculate the area of the triangle with vertices $(1,1,0)$, $(2,1,2)$, $(2,3,3)$.  
We can parametrize the triangle using the function $\Sigma (x,y)=\left (x, y, 2x+\frac{y}{2}-\frac{5}{2}\right )$, right? 
For the boundaries of x and y I have done the following: 
From the verices we see that $1\leq x\leq 2$ and $$0\leq z\leq 3\Rightarrow 0\leq 2x+\frac{y}{2}-\frac{5}{2}\leq 3\Rightarrow \frac{5}{2}-2x\leq \frac{y}{2}\leq 3+\frac{5}{2}-2x\\ \Rightarrow \frac{5}{2}-2x\leq \frac{y}{2}\leq \frac{11}{2}-2x\Rightarrow 5-4x\leq y\leq 11-4x$$ Since from the vertices the smaller value of $y$ is $1$ we get that $1\leq y\leq 11-4x$. 
So, do we calculate the following? 
$$A(\Sigma (D))=\iint_D\|\Sigma_x\times\Sigma_y\|dxdy=\int_1^2\int_1^{11-4x}\frac{\sqrt{21}}{2}dydx$$ 
 [/INST] To find your bounds on $x$ and $y$, project down to the $xy$-plane, where our vertices are $(1,1), (2,1)$ and $(2,3)$. Drawing this triangle helps.

That tells us we should have $x$ going from $1$ to $2$, as you noted, and $y$ should run from $y=1$ up to the line joining $(1,1)$ and $(2,3)$. That's the line $y=2x-1$.
The rest of your set-up appears to be fine.

As a double-check, you should get an area equal to $\frac12\|\langle1,0,2\rangle\times\langle1,2,3\rangle\|$, which is just $\frac12$ the magnitude of the cross-product of two legs of the triangle.
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