The direction cosines of a line are proportional to (2,1,2) and the line intersects a plane perpendicularly at the point (1, -2, 4). What is the distance of the plane from the point (3,2,3) ?

Options

A. `sqrt3`
B. 2
C. `2sqrt2`
D. 4

Explanation

To solve the problem step by step, we will derive the equation of the plane and then calculate the distance from the given point to that plane.

### Step 1: Identify the direction ratios and point of intersection
The direction cosines of the line are proportional to (2, 1, 2). This means the direction ratios (a, b, c) of the line are (2, 1, 2). The line intersects the plane perpendicularly at the point (1, -2, 4).

### Step 2: Write the equation of the plane
The general equation of a plane can be expressed as:
\[ a(x - x_1) + b(y - y_1) + c(z - z_1) = 0 \]
Where (x_1, y_1, z_1) is a point on the plane and (a, b, c) are the direction ratios.

Substituting the values:
- \( a = 2 \)
- \( b = 1 \)
- \( c = 2 \)
- \( (x_1, y_1, z_1) = (1, -2, 4) \)

The equation becomes:
\[ 2(x - 1) + 1(y + 2) + 2(z - 4) = 0 \]

### Step 3: Simplify the equation of the plane
Expanding the equation:
\[ 2x - 2 + y + 2 + 2z - 8 = 0 \]
Combining like terms:
\[ 2x + y + 2z - 8 = 0 \]

### Step 4: Rearranging the equation
Rearranging gives us:
\[ 2x + y + 2z = 8 \]

### Step 5: Calculate the distance from the point (3, 2, 3) to the plane
The formula for the distance \( D \) from a point \( (x_0, y_0, z_0) \) to the plane \( Ax + By + Cz + D = 0 \) is given by:
\[ D = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]

Here, the coefficients from our plane equation \( 2x + y + 2z - 8 = 0 \) are:
- \( A = 2 \)
- \( B = 1 \)
- \( C = 2 \)
- \( D = -8 \)

Substituting the point \( (3, 2, 3) \):
\[ D = \frac{|2(3) + 1(2) + 2(3) - 8|}{\sqrt{2^2 + 1^2 + 2^2}} \]

### Step 6: Calculate the numerator
Calculating the numerator:
\[ 2(3) + 1(2) + 2(3) - 8 = 6 + 2 + 6 - 8 = 6 \]
Thus, the numerator is \( |6| = 6 \).

### Step 7: Calculate the denominator
Calculating the denominator:
\[ \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]

### Step 8: Final calculation of distance
Now substituting back into the distance formula:
\[ D = \frac{6}{3} = 2 \]

### Conclusion
The distance of the plane from the point (3, 2, 3) is **2 units**.

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