A plane passes through a fixed point `(a, b, c)` and direction ratios of the normal to the plane are (2, 3 , 4) find the equation of the plane

Answer

To find the equation of the plane that passes through the point \((a, b, c)\) with direction ratios of the normal vector as \((2, 3, 4)\), we can use the standard form of the equation of a plane.

### Step-by-Step Solution:

1. **Identify the given information**:
- The fixed point through which the plane passes is \((a, b, c)\).
- The direction ratios of the normal to the plane are \((2, 3, 4)\).

2. **Use the general equation of a plane**:
The general equation of a plane can be expressed as:
\[ A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 \]
where \((x_0, y_0, z_0)\) is a point on the plane and \((A, B, C)\) are the direction ratios of the normal to the plane.

3. **Substitute the values**:
Here, \((x_0, y_0, z_0) = (a, b, c)\) and \((A, B, C) = (2, 3, 4)\). Thus, we can substitute these values into the equation:
\[ 2(x - a) + 3(y - b) + 4(z - c) = 0 \]

4. **Expand the equation**:
Expanding the equation gives:
\[ 2x - 2a + 3y - 3b + 4z - 4c = 0 \]

5. **Rearranging the equation**:
We can rearrange the equation to get it into a standard form:
\[ 2x + 3y + 4z - (2a + 3b + 4c) = 0 \]

6. **Final equation of the plane**:
Therefore, the equation of the plane is:
\[ 2x + 3y + 4z = 2a + 3b + 4c \]