A plane passes through `(1, -2, 1)` and is perpendicular to two planes `2x-2y+z=0 and x-y+2z=4.` The distance of the plane from the point (0, 2, 2) is

Options

A. `(3)/(sqrt2)" units"`
B. `4sqrt2" units"`
C. `3sqrt2" units"`
D. `2sqrt2" units"`

Answer

To solve the problem step by step, we need to find the equation of the plane that passes through the point \( (1, -2, 1) \) and is perpendicular to the two given planes. Then we will calculate the distance from the point \( (0, 2, 2) \) to this plane.

### Step 1: Identify the normal vectors of the given planes
The equations of the two planes are:
1. \( 2x - 2y + z = 0 \)
2. \( x - y + 2z = 4 \)

The normal vector of the first plane, \( n_1 \), can be derived from the coefficients of \( x, y, z \):
\[ n_1 = \langle 2, -2, 1 \rangle \]

The normal vector of the second plane, \( n_2 \), is:
\[ n_2 = \langle 1, -1, 2 \rangle \]

### Step 2: Find the normal vector of the required plane
The normal vector \( n_3 \) of the required plane is perpendicular to both \( n_1 \) and \( n_2 \). We can find \( n_3 \) by calculating the cross product \( n_1 \times n_2 \).

\[ n_3 = n_1 \times n_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -2 & 1 \\ 1 & -1 & 2 \end{vmatrix} \]

Calculating the determinant:
\[ n_3 = \mathbf{i}((-2) \cdot 2 - 1 \cdot (-1)) - \mathbf{j}(2 \cdot 2 - 1 \cdot 1) + \mathbf{k}(2 \cdot (-1) - (-2) \cdot 1) \]
\[ = \mathbf{i}(-4 + 1) - \mathbf{j}(4 - 1) + \mathbf{k}(-2 + 2) \]
\[ = \mathbf{i}(-3) - \mathbf{j}(3) + \mathbf{k}(0) \]
\[ = \langle -3, -3, 0 \rangle \]

### Step 3: Write the equation of the plane
The general equation of a plane can be expressed as:
\[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]
where \( (x_0, y_0, z_0) \) is a point on the plane and \( \langle a, b, c \rangle \) is the normal vector.

Using the point \( (1, -2, 1) \) and the normal vector \( n_3 = \langle -3, -3, 0 \rangle \):
\[ -3(x - 1) - 3(y + 2) + 0(z - 1) = 0 \]
Expanding this:
\[ -3x + 3 - 3y - 6 = 0 \]
\[ -3x - 3y - 3 = 0 \]
Dividing through by -3:
\[ x + y + 1 = 0 \]

### Step 4: Calculate the distance from the point \( (0, 2, 2) \) to the plane
The distance \( d \) from a point \( (x_1, y_1, z_1) \) to the plane \( ax + by + cz + d = 0 \) is given by the formula:
\[ d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}} \]

For our plane \( x + y + 1 = 0 \), we have:
- \( a = 1 \)
- \( b = 1 \)
- \( c = 0 \)
- \( d = 1 \)

Substituting the point \( (0, 2, 2) \):
\[ d = \frac{|1(0) + 1(2) + 0(2) + 1|}{\sqrt{1^2 + 1^2 + 0^2}} = \frac{|0 + 2 + 0 + 1|}{\sqrt{1 + 1}} = \frac{|3|}{\sqrt{2}} = \frac{3}{\sqrt{2}} \]

### Final Answer
The distance of the plane from the point \( (0, 2, 2) \) is:
\[ \frac{3}{\sqrt{2}} \]