To find the equation of the plane that passes through the points \( (2, 2, 1) \) and \( (9, 3, 6) \) and is perpendicular to the plane given by the equation \( 2x + 6y + 6z - 1 = 0 \), we can follow these steps:
### Step 1: Identify the normal vector of the given plane
The normal vector of the plane \( 2x + 6y + 6z - 1 = 0 \) can be directly obtained from the coefficients of \( x, y, z \). Thus, the normal vector \( \mathbf{n_1} \) is:
\[
\mathbf{n_1} = (2, 6, 6)
\]
### Step 2: Find the direction vector between the two points
Next, we need to find the direction vector \( \mathbf{d} \) between the points \( (2, 2, 1) \) and \( (9, 3, 6) \):
\[
\mathbf{d} = (9 - 2, 3 - 2, 6 - 1) = (7, 1, 5)
\]
### Step 3: Determine the normal vector of the required plane
Since the required plane is perpendicular to the given plane, its normal vector \( \mathbf{n_2} \) must be perpendicular to \( \mathbf{n_1} \) and also in the direction of \( \mathbf{d} \). Therefore, we can express the normal vector of the required plane as:
\[
\mathbf{n_2} = (A, B, C)
\]
where \( (A, B, C) \) satisfies the condition:
\[
\mathbf{n_1} \cdot \mathbf{n_2} = 0 \quad \text{(perpendicularity condition)}
\]
This gives us the equation:
\[
2A + 6B + 6C = 0 \quad \text{(1)}
\]
### Step 4: Use the point-normal form of the plane equation
The equation of the plane passing through the point \( (2, 2, 1) \) can be written as:
\[
A(x - 2) + B(y - 2) + C(z - 1) = 0
\]
Expanding this, we get:
\[
Ax - 2A + By - 2B + Cz - C = 0 \quad \text{(2)}
\]
### Step 5: Substitute the second point into the plane equation
Substituting the second point \( (9, 3, 6) \) into equation (2):
\[
A(9 - 2) + B(3 - 2) + C(6 - 1) = 0
\]
This simplifies to:
\[
7A + B + 5C = 0 \quad \text{(3)}
\]
### Step 6: Solve the system of equations
Now we have a system of equations:
1. \( 2A + 6B + 6C = 0 \)
2. \( 7A + B + 5C = 0 \)
From equation (1), we can express \( C \) in terms of \( A \) and \( B \):
\[
C = -\frac{1}{3}(2A + 6B) \quad \text{(4)}
\]
Substituting equation (4) into equation (3):
\[
7A + B + 5\left(-\frac{1}{3}(2A + 6B)\right) = 0
\]
Multiplying through by 3 to eliminate the fraction:
\[
21A + 3B - 10A - 30B = 0
\]
This simplifies to:
\[
11A - 27B = 0 \quad \Rightarrow \quad A = \frac{27}{11}B
\]
### Step 7: Substitute back to find \( C \)
Substituting \( A \) back into equation (4):
\[
C = -\frac{1}{3}(2\left(\frac{27}{11}B\right) + 6B) = -\frac{1}{3}\left(\frac{54}{11}B + 6B\right)
\]
Finding a common denominator:
\[
C = -\frac{1}{3}\left(\frac{54 + 66}{11}B\right) = -\frac{120}{33}B
\]
### Step 8: Write the final equation of the plane
Now substituting \( A, B, C \) back into the plane equation, we can choose \( B = 11 \) (for simplicity):
\[
A = 27, \quad B = 11, \quad C = -40
\]
Thus, the equation of the plane becomes:
\[
27(x - 2) + 11(y - 2) - 40(z - 1) = 0
\]
Expanding this gives:
\[
27x + 11y - 40z - 54 - 22 + 40 = 0 \quad \Rightarrow \quad 27x + 11y - 40z - 36 = 0
\]
This can be simplified to:
\[
27x + 11y - 40z = 36
\]
### Final Answer
The equation of the plane is:
\[
27x + 11y - 40z = 36
\]
