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+6z=9`
B. `3x+4y-5z+9=0`
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) that is also perpendicular to the plane given by the equation \(2x + 6y + 6z = 1\), we can follow these steps:

### Step 1: Identify the points and the normal vector of the given plane
We have two points:
- Point A: \( P(2, 2, 1) \)
- Point B: \( Q(9, 3, 6) \)

The normal vector \( \mathbf{N} \) of the given plane \( 2x + 6y + 6z = 1 \) can be directly taken from the coefficients of \( x, y, z \):
\[ \mathbf{N} = (2, 6, 6) \]

### Step 2: Find the vector along the line connecting the two points
To find a vector that lies along the plane, we can calculate the vector \( \overrightarrow{PQ} \) from point \( P \) to point \( Q \):
\[ \overrightarrow{PQ} = Q - P = (9 - 2, 3 - 2, 6 - 1) = (7, 1, 5) \]

### Step 3: Find the normal vector of the required plane
Since the required plane is perpendicular to the given plane, its normal vector \( \mathbf{N'} \) can be found by taking the cross product of \( \overrightarrow{PQ} \) and the normal vector \( \mathbf{N} \):
\[ \mathbf{N'} = \overrightarrow{PQ} \times \mathbf{N} \]

Calculating the cross product:
\[ \mathbf{N'} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 7 & 1 & 5 \\ 2 & 6 & 6 \end{vmatrix} \]

Calculating the determinant:
\[ \mathbf{N'} = \mathbf{i}(1 \cdot 6 - 5 \cdot 6) - \mathbf{j}(7 \cdot 6 - 5 \cdot 2) + \mathbf{k}(7 \cdot 6 - 1 \cdot 2) \]
\[ = \mathbf{i}(6 - 30) - \mathbf{j}(42 - 10) + \mathbf{k}(42 - 2) \]
\[ = -24\mathbf{i} - 32\mathbf{j} + 40\mathbf{k} \]

We can simplify this by taking out \(-8\):
\[ \mathbf{N'} = -8(3\mathbf{i} + 4\mathbf{j} - 5\mathbf{k}) \implies \mathbf{N'} = (3, 4, -5) \]

### Step 4: Use the point-normal form to find the equation of the plane
The equation of a plane can be expressed as:
\[ \mathbf{N'} \cdot (\mathbf{r} - \mathbf{a}) = 0 \]
where \( \mathbf{r} = (x, y, z) \) is a general point on the plane and \( \mathbf{a} = (2, 2, 1) \) is a point on the plane.

Substituting the values:
\[ (3, 4, -5) \cdot ((x, y, z) - (2, 2, 1)) = 0 \]
Expanding this:
\[ 3(x - 2) + 4(y - 2) - 5(z - 1) = 0 \]
\[ 3x - 6 + 4y - 8 - 5z + 5 = 0 \]
\[ 3x + 4y - 5z - 9 = 0 \]

### Final Equation
Thus, the equation of the plane is:
\[ 3x + 4y - 5z = 9 \]