The equation of the plane through the points `(2,2,1) and (9,3,6)` and perpendicular to the plane `2x+6y + 6z=9` is `3x + 4y + z-9=0`

Answer

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 \( 2x + 6y + 6z = 9 \), we will follow these steps:

### Step 1: Find the direction vector of the line through the two points
The direction vector \( \vec{d} \) can be found by subtracting the coordinates of the first point from the second point.

\[ \vec{d} = (9 - 2, 3 - 2, 6 - 1) = (7, 1, 5) \]

### Step 2: Find the normal vector of the given plane
The normal vector \( \vec{n} \) of the plane \( 2x + 6y + 6z = 9 \) can be directly obtained from the coefficients of \( x, y, z \) in the equation.

\[ \vec{n} = (2, 6, 6) \]

### Step 3: Find the normal vector of the required plane
Since the required plane is perpendicular to the given plane, its normal vector \( \vec{N} \) can be found by taking the cross product of the direction vector \( \vec{d} \) and the normal vector \( \vec{n} \).

\[ \vec{N} = \vec{d} \times \vec{n} = (7, 1, 5) \times (2, 6, 6) \]

Calculating the cross product:

\[ \vec{N} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & 1 & 5 \\ 2 & 6 & 6 \end{vmatrix} \]

\[ = \hat{i} (1 \cdot 6 - 5 \cdot 6) - \hat{j} (7 \cdot 6 - 5 \cdot 2) + \hat{k} (7 \cdot 6 - 1 \cdot 2) \]

\[ = \hat{i} (6 - 30) - \hat{j} (42 - 10) + \hat{k} (42 - 2) \]

\[ = \hat{i} (-24) - \hat{j} (32) + \hat{k} (40) \]

Thus,

\[ \vec{N} = (-24, -32, 40) \]

### Step 4: Write the equation of the plane
Using the point-normal form of the equation of a plane, which is given by:

\[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]

where \( (x_0, y_0, z_0) \) is a point on the plane (we can use either of the two points), and \( (a, b, c) \) are the components of the normal vector \( \vec{N} \).

Using point \( (2, 2, 1) \):

\[ -24(x - 2) - 32(y - 2) + 40(z - 1) = 0 \]

Expanding this:

\[ -24x + 48 - 32y + 64 + 40z - 40 = 0 \]

Combining like terms:

\[ -24x - 32y + 40z + 72 = 0 \]

Dividing through by -8 to simplify:

\[ 3x + 4y - 5z - 9 = 0 \]

### Step 5: Rearranging the equation
Rearranging gives us the final equation of the plane:

\[ 3x + 4y + z - 9 = 0 \]

### Final Answer
The equation of the plane is:

\[ 3x + 4y + z - 9 = 0 \]