Equation of plane passing through `(-1,3,2)` and perpendicular to the two planes `x+2y + 2z=5`, and `3x+ 3y + 2z=8`, is

Options

A. `2x-4y + 3z+8=0`
B. `2x+4y + 3z +8=0`
C. `2x+4y - 3z+8=0`
D. `x+y+z=4`

Answer

To find the equation of the plane that passes through the point \((-1, 3, 2)\) and is perpendicular to the two given planes \(x + 2y + 2z = 5\) and \(3x + 3y + 2z = 8\), we can follow these steps:

### Step 1: Identify the normal vectors of the given planes
The normal vector of the first plane \(x + 2y + 2z = 5\) is given by the coefficients of \(x\), \(y\), and \(z\), which is \(\mathbf{n_1} = (1, 2, 2)\).

The normal vector of the second plane \(3x + 3y + 2z = 8\) is \(\mathbf{n_2} = (3, 3, 2)\).

### Step 2: Find the normal vector of the required plane
The required plane is perpendicular to both given planes, so its normal vector \(\mathbf{n}\) can be found using the cross product of \(\mathbf{n_1}\) and \(\mathbf{n_2}\).

\[ \mathbf{n} = \mathbf{n_1} \times \mathbf{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 2 \\ 3 & 3 & 2 \end{vmatrix} \]

Calculating the determinant:

\[ \mathbf{n} = \mathbf{i} \begin{vmatrix} 2 & 2 \\ 3 & 2 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & 2 \\ 3 & 2 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 2 \\ 3 & 3 \end{vmatrix} \]

Calculating each determinant:

1. \(\begin{vmatrix} 2 & 2 \\ 3 & 2 \end{vmatrix} = (2 \cdot 2 - 2 \cdot 3) = 4 - 6 = -2\)
2. \(\begin{vmatrix} 1 & 2 \\ 3 & 2 \end{vmatrix} = (1 \cdot 2 - 2 \cdot 3) = 2 - 6 = -4\)
3. \(\begin{vmatrix} 1 & 2 \\ 3 & 3 \end{vmatrix} = (1 \cdot 3 - 2 \cdot 3) = 3 - 6 = -3\)

Thus, we have:

\[ \mathbf{n} = -2\mathbf{i} + 4\mathbf{j} - 3\mathbf{k} = (-2, 4, -3) \]

### Step 3: Write the equation of the plane
The general equation of a plane is given by:

\[ Ax + By + Cz = D \]

Substituting the normal vector components and the point \((-1, 3, 2)\):

\[ -2(x + 1) + 4(y - 3) - 3(z - 2) = 0 \]

Expanding this:

\[ -2x - 2 + 4y - 12 - 3z + 6 = 0 \]

Combining like terms:

\[ -2x + 4y - 3z - 8 = 0 \]

### Step 4: Rearranging the equation
Multiplying through by -1 to make the coefficients positive:

\[ 2x - 4y + 3z + 8 = 0 \]

### Final Equation
Thus, the equation of the required plane is:

\[ 2x - 4y + 3z + 8 = 0 \]