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To solve the problem, let's proceed step-by-step:

### Step 1: Find the equation of the plane
The given information includes a point \((2, -4, 3)\) through which the plane passes and a normal vector \( \mathbf{n} = (4, 2, 5) \).

The general form of the plane equation is:
\[
a(x - x_0) + b(y - y_0) + c(z - z_0) = 0
\]
where \((x_0, y_0, z_0) = (2, -4, 3)\) is the given point, and \( \mathbf{n} = (a, b, c) = (4, 2, 5) \) is the normal vector.

Substituting the values, we get:
\[
4(x - 2) + 2(y + 4) + 5(z - 3) = 0
\]
Expanding this:
\[
4x - 8 + 2y + 8 + 5z - 15 = 0
\]
\[
4x + 2y + 5z - 15 = 0
\]
Thus, the plane equation is:
\[
4x + 2y + 5z = 15
\]

### Step 2: Find the perpendicular distance from the point \((5, 3, 5)\) to the plane
The formula for the perpendicular distance \(d\) from a point \((x_1, y_1, z_1)\) to the plane \(ax + by + cz + d = 0\) is:
\[
d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}
\]
Substituting \(a = 4\), \(b = 2\), \(c = 5\), \(d = -15\), and the point \((5, 3, 5)\):
\[
d = \frac{|4(5) + 2(3) + 5(5) - 15|}{\sqrt{4^2 + 2^2 + 5^2}}
\]
Calculating the numerator:
\[
= |20 + 6 + 25 - 15| = |36|
\]
For the denominator:
\[
\sqrt{4^2 + 2^2 + 5^2} = \sqrt{16 + 4 + 25} = \sqrt{45} = 3\sqrt{5}
\]
Thus, the distance \(d\) is:
\[
d = \frac{36}{3\sqrt{5}} = \frac{12}{\sqrt{5}} = \frac{12}{5} \sqrt{5}
\]

### Therefore, the correct answer is **(e) \(\frac{12}{5} \sqrt{5}\)**.

Would you like any additional explanations or details on this problem? Here are five related questions for further exploration:

1. How do you derive the equation of a plane given two points and a normal vector?
2. What is the significance of the normal vector in the equation of a plane?
3. How do you determine the shortest distance between two parallel planes?
4. What is the general formula for finding the distance between a point and a plane?
5. How does this problem change if the given point lies on the plane itself?

**Tip:** Remember, finding the perpendicular distance to a plane always involves the normal vector, as it represents the shortest path.

Carilah persamaan bidang yang melalui titik (2, -4, 3) dan tegak lurus dengan vektor n = (4, 2, 5). Selanjutnya carilah jarak tegak lurus dari titik (5, 3, 5) ke bidang tersebut.

Learn how to find the equation of a plane passing through point (2, -4, 3) with normal vector n = (4, 2, 5), and calculate the perpendicular distance from point (5, 3, 5) to the plane. This problem involves key concepts from analytic geometry and vector mathematics.

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