Let $D$ be a closed disk (w/ boundary $C$) and let $D_a$, $D_b$ be two disjoint closed disks in the interior of $D$ (w/ boundaries $C_a$ and $C_b$, resp.) . Now remove the interiors of $D_a$ and $D_b$, and identify the points on $C_a$ that are 90 degrees apart, the points on $C_b$ that are 120 degrees apart, and the points on $C$ that are 180 degrees apart (antipodal identification).
What is the fundamental group of this resulting space?
I know that the fundamental group depends only on the 2-skeleton, and that if I have for example a circle and I attach a 2-cell via a map of degree $m$ then I will end up w/ a 2-complex w/ fundamental group $\mathbb{Z_m}$. And if I attach another 2-cell via a map of degree $n$ then I will end up w/ a 2-complex w/ fundamental group $\mathbb{Z_{(m,n)}}$. I am having trouble applying this or a similar type of reasoning here though...
Another way to look at it is to start with the wedge sum of three circles, each representing one of $C$, $C_a$, $C_b$. Let $\pi_1(S^1 \vee S^1 \vee S^1) = \langle x, y, z \rangle$, the free group on three generators. The space you have results from this wedge sum by attaching a $2$-cell along $x^2y^3z^4$. Therefore, the fundamental group of the given space is $$ \langle x, y, z \mid x^2y^3z^4 \rangle. $$
Of course, you can still apply Van Kampen. It will give you the same result as long as you handle the inclusion map right. When the generator of the fundamental group of intersection is mapped into each subspace, it's mapped to multiples of generators according to identification.