My attempt:
Consider $AG(2,n)$ with $n\geq 3$. We know that $AG(2,n)$ is a BIBD with parameters $(n^2,n^2+n, n+1,n,1)$ $\implies$ $b=n^2+n$. But for any t-design, $b=$ $\lambda {v\choose t}\over {k \choose t}$ = $(1) {n^2\choose 3} \over {n \choose 3}$ = $\frac{n(n^2-2)(n^2-1)}{(n-2)(n-1)}.$ Now, $n^2+n = \frac{n(n^2-2)(n^2-1)}{(n-2)(n-1)}$ $\implies$ $n=-1$ or $n=0$. But we said $n\geq3$, which implies that an $AG(2,n)$ is never a 3-design.
So I have a couple of concerns. Firstly, is what I did correct at all? I feel nervous about equating both the expressions for $b$ and reasoning from that. Secondly, I don't understand how we get a "trivial" 3-design for $n=2$. What does that mean?
Any assistance is much appreciated!
Your argument is almost correct, except for the fact that you prematurely set $\lambda=1$; a general $3$-design may have different values for $\lambda$.
If you leave the $\lambda$ when equating the expressions for $b$, you can reduce your equation to $$n-2=\lambda\cdot(n^2-2)$$ which has the unique integral solution $(n,\lambda)=(2,0)$, and this is in fact what's meant by "trivial" solution.