In $\mathbb{RP}^4$ let $\pi_1$ and $\pi_2$ be planes and $l$ a line, given by \begin{eqnarray*} π_1:&\quad x+3z-s=0,&\quad 2x+3y+t=0,\\ π_2:&\quad -x+z+2t=0,&\quad 3x+y=0,\\ l:&\quad -13x+3z=0,&\quad 7y+3t=0,\quad -38y+3s=0. \end{eqnarray*} I want to determine the planes $π$ that contain $l$ and intersect with each $π_1$ and $π_2$ in a line.
I already found that $π_1 \cap π_2 = (-1:3:13:-7:38)$ and that $l$ does not intersect with either one of the two planes.
Unfortunately i am stuck with the further procedure and thus appreciate any help very much!
If a plane $\pi$ intersects the plane $\pi_i$ in a line $m$ and contains the line $l$, then $\pi$ contains both $m$ and $l$ and hence the two lines intersect. In particular $l$ intersects $\pi_i$, which you have shown not to be the case, hence no such plane $\pi$ exists.