Find the equation of the plane passing through the points A(1,-1,2) and B(2,-2,2) and perpendicular to the plane `6x-2y+2z=9`.

Answer

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

### Step 1: Find the direction vector AB
The direction vector \( \vec{AB} \) can be calculated using the coordinates of points A and B.

\[ \vec{AB} = B - A = (2 - 1) \hat{i} + (-2 + 1) \hat{j} + (2 - 2) \hat{k} = 1 \hat{i} - 1 \hat{j} + 0 \hat{k} \]

Thus,

\[ \vec{AB} = \langle 1, -1, 0 \rangle \]

**Hint:** To find the direction vector between two points, subtract the coordinates of the first point from the second point.

### Step 2: Identify the normal vector of the given plane
The normal vector \( \vec{n_1} \) of the plane \( 6x - 2y + 2z = 9 \) can be directly taken from the coefficients of \(x\), \(y\), and \(z\).

\[ \vec{n_1} = \langle 6, -2, 2 \rangle \]

**Hint:** The normal vector of a plane can be found from the coefficients of \(x\), \(y\), and \(z\) in the plane's equation.

### Step 3: Find the normal vector of the required plane
The normal vector \( \vec{n} \) of the required plane must be perpendicular to both \( \vec{AB} \) and \( \vec{n_1} \). We can find this normal vector by taking the cross product \( \vec{AB} \times \vec{n_1} \).

\[ \vec{n} = \vec{AB} \times \vec{n_1} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 0 \\ 6 & -2 & 2 \end{vmatrix} \]

Calculating the determinant:

\[ \vec{n} = \hat{i} \begin{vmatrix} -1 & 0 \\ -2 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 0 \\ 6 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -1 \\ 6 & -2 \end{vmatrix} \]

Calculating each of these determinants:

1. \( \hat{i}((-1)(2) - (0)(-2)) = -2 \hat{i} \)
2. \( -\hat{j}((1)(2) - (0)(6)) = -2 \hat{j} \)
3. \( \hat{k}((1)(-2) - (-1)(6)) = -2 + 6 = 4 \hat{k} \)

So,

\[ \vec{n} = \langle -2, -2, 4 \rangle \]

**Hint:** The cross product of two vectors gives a vector that is perpendicular to both.

### Step 4: Write the equation of the plane
The equation of a plane can be expressed using the normal vector \( \vec{n} = \langle A, B, C \rangle \) and a point \( (x_0, y_0, z_0) \) on the plane:

\[ A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 \]

Using point A(1, -1, 2) and the normal vector \( \vec{n} = \langle -2, -2, 4 \rangle \):

\[ -2(x - 1) - 2(y + 1) + 4(z - 2) = 0 \]

Expanding this equation:

\[ -2x + 2 - 2y - 2 + 4z - 8 = 0 \]

Combining like terms:

\[ -2x - 2y + 4z - 8 = 0 \]

Dividing through by -2 gives:

\[ x + y - 2z + 4 = 0 \]

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

\[ x + y - 2z + 4 = 0 \]