Let $L_1$ and $L_2$ $\in$ $\mathbb{P}^4$ be two planes that intersect in exactly one point $Q$. Let $P_1 \in L_1$, $P_2 \in L_2$ points, such that $P_1 \neq Q \neq P_2$. Using the duality theorem, show that there is exactly one plane in $\mathbb{P}^4$ that goes through $P_1$ and $P_2$ and intersects $L_1$ and $L_2$ in a line. Determine this plane.

The duality theorem states the following: Let $F^1, F^2$ be two transversal flags, and let $\lambda, \mu$ be two acceptable partitions with respect to $G(d, n)$, such that $|\lambda|+ |\mu| = (n-d) \cdot d$. Then: (1) If $\lambda, \mu$ are complementary, then $\Omega (F^1, \lambda) \cap \Omega( F^2, \mu)= 1$ point. (2) Otherwise, $\Omega( F^1, \lambda) \cap \Omega( F^2, \mu)= \emptyset$.

The way I wanted to solve this problem was by finding all Schubert classes of G(3,5). There are 9 acceptable partitions (inclusive the zero partition), and calculating their intersection after choosing the right Schubert classes.

However, in order to choose the right Schubert classes, I need to be able to interpret them geometrically, and that is where I am stuck at.

I would appreciate your help a lot. Thank you in advance.