Find the foot of perpendicular drawn from the point P(0, 0, 0) to the plane `x + 2y + 2z =13.`

Answer

To find the foot of the perpendicular drawn from the point \( P(0, 0, 0) \) to the plane given by the equation \( x + 2y + 2z = 13 \), we can use the formula for the foot of the perpendicular from a point to a plane.

### Step-by-Step Solution:

1. **Identify the Plane Equation and Point**:
The plane is given by the equation:
\[ x + 2y + 2z = 13 \]
The point \( P \) is given as \( (0, 0, 0) \).

2. **Extract Coefficients from the Plane Equation**:
From the plane equation \( Ax + By + Cz = D \), we can identify:
- \( A = 1 \)
- \( B = 2 \)
- \( C = 2 \)
- \( D = 13 \)

3. **Substitute the Point Coordinates**:
Let the coordinates of the foot of the perpendicular be \( (x, y, z) \). According to the formula:
\[ \frac{x - x_1}{A} = \frac{y - y_1}{B} = \frac{z - z_1}{C} = \frac{-Ax_1 - By_1 - Cz_1 + D}{\sqrt{A^2 + B^2 + C^2}} \]
Here, \( (x_1, y_1, z_1) = (0, 0, 0) \).

4. **Set Up the Equation**:
The equation simplifies to:
\[ \frac{x - 0}{1} = \frac{y - 0}{2} = \frac{z - 0}{2} = \frac{-1(0) - 2(0) - 2(0) + 13}{\sqrt{1^2 + 2^2 + 2^2}} \]
This further simplifies to:
\[ \frac{x}{1} = \frac{y}{2} = \frac{z}{2} = \frac{13}{\sqrt{1 + 4 + 4}} \]

5. **Calculate the Denominator**:
Calculate \( \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \).

6. **Final Equation**:
Now we have:
\[ \frac{x}{1} = \frac{y}{2} = \frac{z}{2} = \frac{13}{3} \]

7. **Solve for \( x, y, z \)**:
From this, we can express \( x, y, z \) as:
- \( x = \frac{13}{3} \)
- \( y = 2 \cdot \frac{13}{3} = \frac{26}{3} \)
- \( z = 2 \cdot \frac{13}{3} = \frac{26}{3} \)

8. **Foot of the Perpendicular**:
Thus, the coordinates of the foot of the perpendicular \( F \) are:
\[ F\left(\frac{13}{3}, \frac{26}{3}, \frac{26}{3}\right) \]

### Final Answer:
The foot of the perpendicular from the point \( P(0, 0, 0) \) to the plane \( x + 2y + 2z = 13 \) is:
\[ F\left(\frac{13}{3}, \frac{26}{3}, \frac{26}{3}\right) \]