Step 1 (i): Gauss's Law: the total electric flux through any closed surface equals $q_{enc}/\varepsilon_0$.

Step 2: For an infinite charged sheet with surface charge density $\sigma$, take a cylindrical (pillbox) Gaussian surface straddling the sheet, with flat faces of area $A$ parallel to the sheet. By symmetry, field is perpendicular to the sheet, same magnitude on both sides.

Step 3: Flux $=2EA$ (through both faces; no flux through curved surface). Enclosed charge $=\sigma A$.

Step 4: $2EA = \sigma A/\varepsilon_0 \Rightarrow E = \dfrac{\sigma}{2\varepsilon_0}$, directed away from the sheet (for positive $\sigma$).

Step 5 (ii): Field due to wire 1 (line charge) at P (10 cm away): $E_1 = \dfrac{\lambda_1}{2\pi\varepsilon_0(0.10)} = \dfrac{2k\lambda_1}{0.10} = \dfrac{2\times9\times10^9\times10\times10^{-6}}{0.10} = 1.8\times10^6$ N/C, directed away from wire 1 (since $\lambda_1>0$).

Step 6: Field due to wire 2 (20 cm away): $E_2 = \dfrac{2k\lambda_2}{0.20} = \dfrac{2\times9\times10^9\times(-20\times10^{-6})}{0.20} = -1.8\times10^6$ N/C in magnitude direction — since $\lambda_2$ is negative, field points TOWARD wire 2; magnitude $=1.8\times10^6$ N/C.

Step 7: Both fields point in the same direction at P (away from wire1 = toward wire2, since wire1 pushes away and wire2's negative charge attracts toward it — both effects point the same way, from wire1 toward wire2). So $E_{net} = E_1+E_2 = 1.8\times10^6+1.8\times10^6=3.6\times10^6$ N/C.

Step 8: Force on electron: $F = eE_{net} = 1.6\times10^{-19}\times3.6\times10^6 = 5.76\times10^{-13}$ N, directed opposite to $E$ (since electron is negative) — i.e., from wire 2 toward wire 1.