The electric flux is given by:

\[
\Phi = \overrightarrow{E} \cdot \overrightarrow{S}
\]

Here, \(\overrightarrow{E} = 2\hat{i} + 4\hat{j} + 3\hat{k}\) and \(\overrightarrow{S} = 10\hat{j}\).

\[
\Phi = (2\hat{i} + 4\hat{j} + 3\hat{k}) \cdot (10\hat{j}) = 0 + 4 \cdot 10 + 0 = 40 \, \text{units}.
\]

Step-by-Step Explanation:

1. Flux through the Entire Pyramid:
The total charge enclosed by the pyramid is \(+Q\), and the total flux through the entire closed surface of the pyramid is given by Gauss's law:
\[
\Phi_{\text{total}} = \frac{Q}{\varepsilon_0}.
\]

2. Flux Distribution:
The pyramid has a square base and four triangular faces. However, **the charge \(+Q\) is located at the center of the base, not at the geometric center of the pyramid.** This means the flux through the base is not zero and contributes to the total flux.

3. Flux Through the Base:
Due to symmetry, half of the total flux passes through the square base:
\[
\Phi_{\text{base}} = \frac{\Phi_{\text{total}}}{2} = \frac{Q}{2\varepsilon_0}.
\]

4. Flux Through the Four Triangular Faces:
The remaining half of the total flux passes through the four triangular faces combined:
\[
\Phi_{\text{triangular (total)}} = \frac{\Phi_{\text{total}}}{2} = \frac{Q}{2\varepsilon_0}.
\]

5. Flux Through One Triangular Face:
Since the four triangular faces are identical, the flux is equally distributed among them:
\[
\Phi_{\text{face}} = \frac{\Phi_{\text{triangular (total)}}}{4} = \frac{\frac{Q}{2\varepsilon_0}}{4} = \frac{Q}{8\varepsilon_0}.
\]

Final Answer:
The flux through one triangular face is:
\[
{\frac{Q}{8\varepsilon_0}}.
\]

Since the electric field \( E \) is parallel to the cylinder's axis, the field lines enter and exit symmetrically through the two flat circular ends. The curved surface has no perpendicular component of the field, so no flux passes through it.

By Gauss's law, the net flux through the entire surface is:
\[
\Phi = \text{Charge enclosed} / \varepsilon_0
\]

As no charge is enclosed, \( \Phi = 0 \).

The electric flux through the hemisphere is due to the uniform electric field passing perpendicularly through the circular plane of radius \( R \).

Flux through the circular plane:
\[
\Phi = E \cdot \text{Area of circular plane} = E \cdot \pi R^2
\]

Since no flux passes through the curved surface (field is parallel to it), the total flux is:
\[
\Phi = \pi R^2 E
\]

Gauss's law states that the total electric flux passing through a closed surface is equal to the charge enclosed within the surface divided by the permittivity of the medium.

In short: The net electric field around a closed surface depends on the charge inside it.