What is the distance of the point (2,3,4) from the plane 3x-6y + 2z + 11 = 0

Options

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

Answer

To find the distance of the point \( P(2, 3, 4) \) 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 the plane \( Ax + By + Cz + D = 0 \):

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

### Step-by-Step Solution:

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

2. **Substitute the coordinates of the point \( P(2, 3, 4) \):**
- Here, \( x_1 = 2 \), \( y_1 = 3 \), and \( z_1 = 4 \).

3. **Calculate \( Ax_1 + By_1 + Cz_1 + D \):**
- Calculate each term:
- \( Ax_1 = 3 \cdot 2 = 6 \)
- \( By_1 = -6 \cdot 3 = -18 \)
- \( Cz_1 = 2 \cdot 4 = 8 \)
- Now, sum these with \( D \):
\[ Ax_1 + By_1 + Cz_1 + D = 6 - 18 + 8 + 11 = 7 \]

4. **Calculate the absolute value:**
- \( |Ax_1 + By_1 + Cz_1 + D| = |7| = 7 \)

5. **Calculate the denominator \( \sqrt{A^2 + B^2 + C^2} \):**
- Calculate each square:
- \( A^2 = 3^2 = 9 \)
- \( B^2 = (-6)^2 = 36 \)
- \( C^2 = 2^2 = 4 \)
- Now, sum these:
\[ A^2 + B^2 + C^2 = 9 + 36 + 4 = 49 \]
- Take the square root:
\[ \sqrt{49} = 7 \]

6. **Calculate the distance \( d \):**
\[ d = \frac{7}{7} = 1 \]

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