What is the equation to the planes through (1,2,3) parallel to `3x+4y-5z=0`?

Options

A. `3x+4y+5z+4=0`
B. `3x+4y-5z+14=0`
C. `3x+4y-5z+4=0`
D. `3x+4y-5z-4=0`

Explanation

To find the equation of the plane that passes through the point (1, 2, 3) and is parallel to the plane given by the equation \(3x + 4y - 5z = 0\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Normal Vector**:
The normal vector of the given plane \(3x + 4y - 5z = 0\) can be derived from the coefficients of \(x\), \(y\), and \(z\). Thus, the normal vector \(\vec{n}\) is \((3, 4, -5)\).

**Hint**: The normal vector of a plane can be found directly from the coefficients of \(x\), \(y\), and \(z\) in the plane's equation.

2. **Use the Point-Normal Form of the Plane**:
The equation of a plane in point-normal form is given by:
\[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]
where \((x_0, y_0, z_0)\) is a point on the plane and \((a, b, c)\) are the components of the normal vector. Here, we have the point \((1, 2, 3)\) and the normal vector \((3, 4, -5)\).

3. **Substitute the Values**:
Substituting the values into the point-normal form:
\[ 3(x - 1) + 4(y - 2) - 5(z - 3) = 0 \]

4. **Expand the Equation**:
Expanding the equation gives:
\[ 3x - 3 + 4y - 8 - 5z + 15 = 0 \]
Simplifying this, we get:
\[ 3x + 4y - 5z + 4 = 0 \]

5. **Rearranging the Equation**:
To express it in the standard form, we can rewrite it as:
\[ 3x + 4y - 5z = -4 \]

### Final Answer:
The equation of the plane that passes through the point \((1, 2, 3)\) and is parallel to the plane \(3x + 4y - 5z = 0\) is:
\[ 3x + 4y - 5z = -4 \]