The distance of the point `(1,-2,4)` from the plane passing through the point `(1,2,2)` perpendicular to the planes `x-y+2z=3` and `2x-2y+z+12=0` is

Options

A. `2`
B. `sqrt2`
C. `2sqrt2`
D. `(1)/(sqrt2)`

Answer

To find the distance of the point \( P(1, -2, 4) \) from the plane passing through the point \( A(1, 2, 2) \) and perpendicular to the two given planes, we will follow these steps:

### Step 1: Find the normals of the given planes
The equations of the given planes are:
1. \( x - y + 2z = 3 \)
2. \( 2x - 2y + z + 12 = 0 \)

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

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

### Step 2: Find the normal of the required plane
To find the normal vector of the required plane, we take the cross product of \( n_1 \) and \( n_2 \):
\[ n = n_1 \times n_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 2 & -2 & 1 \end{vmatrix} \]

Calculating the determinant:
\[ n = \hat{i}((-1)(1) - (2)(-2)) - \hat{j}((1)(1) - (2)(2)) + \hat{k}((1)(-2) - (-1)(2)) \]
\[ = \hat{i}(-1 + 4) - \hat{j}(1 - 4) + \hat{k}(-2 + 2) \]
\[ = 3\hat{i} + 3\hat{j} + 0\hat{k} \]
Thus, the normal vector of the required plane is:
\[ n = (3, 3, 0) \]

### Step 3: Find the equation of the required plane
Using the point \( A(1, 2, 2) \) and the normal vector \( n(3, 3, 0) \), we can write the equation of the plane in the form:
\[ n \cdot (r - a) = 0 \]
Where \( r = (x, y, z) \) and \( a = (1, 2, 2) \):
\[ 3(x - 1) + 3(y - 2) + 0(z - 2) = 0 \]
Simplifying this:
\[ 3x - 3 + 3y - 6 = 0 \implies 3x + 3y = 9 \implies x + y = 3 \]

### Step 4: Find the distance from point \( P(1, -2, 4) \) to the plane
To find the distance \( d \) from point \( P(1, -2, 4) \) to the plane \( x + y = 3 \), we can use the formula for the distance from a point to a plane:
\[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
For the plane \( x + y - 3 = 0 \):
- \( A = 1, B = 1, C = 0, D = -3 \)
- The point \( P(1, -2, 4) \) gives \( x_0 = 1, y_0 = -2, z_0 = 4 \)

Substituting these values:
\[ d = \frac{|1(1) + 1(-2) + 0(4) - 3|}{\sqrt{1^2 + 1^2 + 0^2}} = \frac{|1 - 2 - 3|}{\sqrt{2}} = \frac{|-4|}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \]

### Final Answer
The distance of the point \( (1, -2, 4) \) from the plane is \( 2\sqrt{2} \) units.