Which one of the following is the plane containing the line `(x-2)/2=(y-3)/3=(z-4)/5` and parallel to z-axis?

Options

A. `2x-3y=0`
B. `5x-2z=0`
C. `5y-3z=0`
D. `3x-2y=0`

Explanation

To solve the problem, we need to find the equation of the plane that contains the given line and is parallel to the z-axis.

### Step-by-Step Solution:

1. **Identify the line's parametric equations**:
The line is given in the symmetric form:
\[ \frac{x-2}{2} = \frac{y-3}{3} = \frac{z-4}{5} \]
Let \( r \) be the parameter. We can express \( x, y, z \) in terms of \( r \):
\[ x = 2r + 2 \]
\[ y = 3r + 3 \]
\[ z = 5r + 4 \]

2. **Determine the point on the line**:
Since the plane is parallel to the z-axis, we can set \( z = 0 \) to find a specific point on the plane. Setting \( z = 0 \):
\[ 0 = 5r + 4 \]
Solving for \( r \):
\[ 5r = -4 \implies r = -\frac{4}{5} \]

3. **Substitute \( r \) back to find \( x \) and \( y \)**:
Now substitute \( r = -\frac{4}{5} \) into the equations for \( x \) and \( y \):
\[ x = 2\left(-\frac{4}{5}\right) + 2 = -\frac{8}{5} + 2 = -\frac{8}{5} + \frac{10}{5} = \frac{2}{5} \]
\[ y = 3\left(-\frac{4}{5}\right) + 3 = -\frac{12}{5} + 3 = -\frac{12}{5} + \frac{15}{5} = \frac{3}{5} \]

4. **Formulate the equation of the plane**:
The plane is parallel to the z-axis, which means it can be represented as a linear equation in \( x \) and \( y \). Since we have the point \( \left( \frac{2}{5}, \frac{3}{5}, 0 \right) \), we need to find a relationship between \( x \) and \( y \).
We can express this relationship as:
\[ 3x - 2y = 0 \]
This is derived from the cross-multiplication of the ratios we found earlier.

5. **Final equation of the plane**:
The equation of the plane can be written as:
\[ 3x - 2y = 0 \quad \text{and} \quad z = 0 \]

Thus, the equation of the required plane is:
\[ 3x - 2y = 0 \quad \text{(and parallel to the z-axis)} \]