The problem of point charge between two conducting planes is more easily tackled (if we want only the total charge induced on the planes) if we replace the point charge by a uniformly charged plane sheet. Let $\sigma$ be the charge density on this sheet and $E_{1}, E_{2}$ outward electric field on the two sides of this sheet.
Then $\quad E_{1}+E_{2}=\frac{\sigma}{\varepsilon_{0}}$
The conducting planes will be assumed to be grounded. Then $E_{1} x=E_{2}(l-x)$.
Hence
$$
E_{1}=\frac{\sigma}{l \varepsilon_{0}}(l-x), E_{2}=\frac{\sigma}{l \varepsilon_{0}} x
$$
This means that the induced charge density on the plane conductors are
$$
\sigma_{1}=-\frac{\sigma}{l}(l-x), \sigma_{2}=-\frac{\sigma}{l} x
$$
Hence $q_{1}=-\frac{q}{l}(l-x), q_{2}=-\frac{q}{l} x$