The equation of the plane through the point `(2,5,-3)` perpendicular to the planes `x + 2y + 2z =1 and x-2y + 2z =4` is :

Options

A. `3x -4y + 2z -20=0`
B. `7x -y + 5x =30`
C. `2x- z=7`
D. ` 10x-y -4z =27`

Answer

To find the equation of the plane that passes through the point \( (2, 5, -3) \) and is perpendicular to the given planes \( x + 2y + 2z = 1 \) and \( x - 2y + 2z = 4 \), we will follow these steps:

### Step 1: Identify the normal vectors of the given planes
The normal vector of a plane given by the equation \( Ax + By + Cz = D \) is \( \vec{n} = (A, B, C) \).

For the first plane \( P_1: x + 2y + 2z = 1 \):
- The normal vector \( \vec{n_1} = (1, 2, 2) \).

For the second plane \( P_2: x - 2y + 2z = 4 \):
- The normal vector \( \vec{n_2} = (1, -2, 2) \).

### Step 2: Find the normal vector of the required plane
The normal vector of the required plane will be perpendicular to both \( \vec{n_1} \) and \( \vec{n_2} \). We can find this normal vector by calculating the cross product \( \vec{n_1} \times \vec{n_2} \).

\[ \vec{n_1} = (1, 2, 2), \quad \vec{n_2} = (1, -2, 2) \]

Calculating the cross product:

\[ \vec{n} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 1 & -2 & 2 \end{vmatrix} \]

Calculating the determinant:

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

Calculating each of the minors:

\[ \vec{n} = \hat{i} (2 \cdot 2 - 2 \cdot (-2)) - \hat{j} (1 \cdot 2 - 1 \cdot 2) + \hat{k} (1 \cdot (-2) - 1 \cdot 2) \]

\[ = \hat{i} (4 + 4) - \hat{j} (2 - 2) + \hat{k} (-2 - 2) \]

\[ = \hat{i} (8) - \hat{j} (0) - \hat{k} (4) \]

Thus, the normal vector \( \vec{n} = (8, 0, -4) \).

### Step 3: Write the equation of the plane
The general equation of a plane with normal vector \( (A, B, C) \) passing through the point \( (x_0, y_0, z_0) \) is given by:

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

Substituting \( A = 8, B = 0, C = -4 \) and the point \( (2, 5, -3) \):

\[ 8(x - 2) + 0(y - 5) - 4(z + 3) = 0 \]

Expanding this:

\[ 8x - 16 - 4z - 12 = 0 \]

Combining like terms:

\[ 8x - 4z - 28 = 0 \]

Rearranging gives:

\[ 8x - 4z = 28 \]

Dividing through by 4:

\[ 2x - z = 7 \]

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

\[ 2x - z = 7 \]