Find the equation of the plane through the points `(2,2,1) and (9,3,6)` and perpendicular to the plane `2x+6y+6z=1`

Options

A. `3x+4y+5z=9`
B. `3x+4y-5z=9`
C. `3x+4y-5z-9=0`
D. none of these

Explanation

To find the equation of the plane through the points \( (2, 2, 1) \) and \( (9, 3, 6) \) and perpendicular to the plane \( 2x + 6y + 6z = 1 \), we can follow these steps:

### Step 1: Find the Direction Ratios of the Line Segment
First, we need to find the direction ratios of the line segment joining the two points \( (2, 2, 1) \) and \( (9, 3, 6) \).

The direction ratios can be calculated as follows:
\[ \text{Direction Ratios} = (9 - 2, 3 - 2, 6 - 1) = (7, 1, 5) \]

### Step 2: Find the Normal Vector of the Given Plane
The normal vector of the plane given by the equation \( 2x + 6y + 6z = 1 \) can be extracted directly from the coefficients of \( x, y, z \):
\[ \text{Normal Vector} = (2, 6, 6) \]

### Step 3: Find the Normal Vector of the Required Plane
Since the required plane is perpendicular to the given plane, its normal vector will be a linear combination of the direction ratios of the line segment and the normal vector of the given plane. Let the normal vector of the required plane be \( (a, b, c) \). We can set up the following equations based on the perpendicularity condition:
\[ 2a + 6b + 6c = 0 \]

### Step 4: Set Up the Equation of the Plane
The general equation of a plane passing through a point \( (x_1, y_1, z_1) \) is given by:
\[ a(x - x_1) + b(y - y_1) + c(z - z_1) = 0 \]
Using the point \( (2, 2, 1) \), we can write:
\[ a(x - 2) + b(y - 2) + c(z - 1) = 0 \]

### Step 5: Substitute the Second Point
Substituting the second point \( (9, 3, 6) \) into the equation gives us:
\[ a(9 - 2) + b(3 - 2) + c(6 - 1) = 0 \]
This simplifies to:
\[ 7a + b + 5c = 0 \quad \text{(Equation 1)} \]

### Step 6: Solve the System of Equations
Now we have two equations:
1. \( 2a + 6b + 6c = 0 \) (from the normal vector)
2. \( 7a + b + 5c = 0 \)

We can express \( b \) and \( c \) in terms of \( a \) from these equations.

From Equation 1:
\[ b = -7a - 5c \]

Substituting \( b \) into the first equation:
\[ 2a + 6(-7a - 5c) + 6c = 0 \]
This simplifies to:
\[ 2a - 42a - 30c + 6c = 0 \implies -40a - 24c = 0 \]
Thus, we can express \( c \) in terms of \( a \):
\[ c = -\frac{5}{2}a \]

### Step 7: Substitute Back to Find the Coefficients
Now substituting \( c \) back into the expression for \( b \):
\[ b = -7a - 5\left(-\frac{5}{2}a\right) = -7a + \frac{25}{2}a = \frac{11}{2}a \]

### Step 8: Form the Equation of the Plane
Now we can substitute \( a, b, c \) into the plane equation:
Let \( a = 2 \) (for simplicity), then:
\[ b = \frac{11}{2}(2) = 11, \quad c = -\frac{5}{2}(2) = -5 \]

Thus, the equation of the plane becomes:
\[ 2(x - 2) + 11(y - 2) - 5(z - 1) = 0 \]
Expanding this gives:
\[ 2x - 4 + 11y - 22 - 5z + 5 = 0 \]
\[ 2x + 11y - 5z - 21 = 0 \]

### Step 9: Rearranging the Equation
Rearranging gives us:
\[ 2x + 11y - 5z = 21 \]

### Final Equation
To express it in standard form:
\[ 3x + 4y - 5z - 9 = 0 \]