What is the distance of the point (2,3,4) from the plane `3x-6y+2z+11=?`

Options

A. 1 unit
B. 2 unit
C. 3 unit
D. 4 units

Explanation

To find the distance of the point \( (2, 3, 4) \) from the plane given by the equation \( 3x - 6y + 2z + 11 = 0 \), we can use the formula for the distance \( D \) from a point \( (x_1, y_1, z_1) \) to a plane given by the equation \( Ax + By + Cz + D = 0 \):

\[ D = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \]

### Step-by-Step Solution:

1. **Identify the coefficients from the plane equation**:
The plane equation is \( 3x - 6y + 2z + 11 = 0 \). Here, we can identify:
- \( A = 3 \)
- \( B = -6 \)
- \( C = 2 \)
- \( D = 11 \)

2. **Substitute the point coordinates into the formula**:
The point given is \( (x_1, y_1, z_1) = (2, 3, 4) \). We substitute these values into the formula:
\[ D = \frac{|3(2) + (-6)(3) + 2(4) + 11|}{\sqrt{3^2 + (-6)^2 + 2^2}} \]

3. **Calculate the numerator**:
First, calculate \( 3(2) + (-6)(3) + 2(4) + 11 \):
\[ = 6 - 18 + 8 + 11 \]
\[ = 6 - 18 + 8 + 11 = 7 \]
Thus, the absolute value is \( |7| = 7 \).

4. **Calculate the denominator**:
Now, calculate \( \sqrt{3^2 + (-6)^2 + 2^2} \):
\[ = \sqrt{9 + 36 + 4} = \sqrt{49} = 7 \]

5. **Combine the results**:
Now substitute back into the distance formula:
\[ D = \frac{7}{7} = 1 \]

### Final Answer:
The distance of the point \( (2, 3, 4) \) from the plane \( 3x - 6y + 2z + 11 = 0 \) is \( 1 \) unit.