For all `d,0ltdlt1`, which one of the following points is the reflection of the point `(d,2d,3d)` in the plane pasing through the points `(1,0,0),(0,1,0` and `(0,0,1)`?

Options

A. `(2/3-3d,2/3-2d,2/3-d)`
B. `(-1/3+3d,2d,1/3+e)`
C. `(3d,2d,d)`
D. `(1/3+d,2/3-2d,-1/3+d)`

Answer

To find the reflection of the point \( P(d, 2d, 3d) \) in the plane defined by the points \( (1, 0, 0) \), \( (0, 1, 0) \), and \( (0, 0, 1) \), we will follow these steps:

### Step 1: Find the equation of the plane
The plane can be determined using the normal vector derived from the cross product of two vectors formed by the given points.

1. **Vectors in the plane**:
- \( \vec{A} = (0, 1, 0) - (1, 0, 0) = (-1, 1, 0) \)
- \( \vec{B} = (0, 0, 1) - (1, 0, 0) = (-1, 0, 1) \)

2. **Cross product**:
\[ \vec{N} = \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{vmatrix} = (1 \cdot 1 - 0 \cdot 0) \hat{i} - (0 \cdot 1 - (-1) \cdot 1) \hat{j} + ((-1) \cdot 0 - (-1) \cdot 1) \hat{k} \]
\[ = (1) \hat{i} - (1) \hat{j} + (1) \hat{k} = (1, -1, 1) \]

3. **Equation of the plane**:
Using the normal vector \( (1, -1, 1) \) and point \( (1, 0, 0) \):
\[ 1(x - 1) - 1(y - 0) + 1(z - 0) = 0 \]
Simplifying gives:
\[ x - y + z - 1 = 0 \quad \text{or} \quad x - y + z = 1 \]

### Step 2: Find the foot of the perpendicular from point \( P \) to the plane
Let the foot of the perpendicular from point \( P(d, 2d, 3d) \) to the plane be \( N(x_1, y_1, z_1) \).

1. **Direction ratios**:
The direction ratios of the normal are \( (1, -1, 1) \). Thus, we can write the parametric equations:
\[ \frac{x - d}{1} = \frac{y - 2d}{-1} = \frac{z - 3d}{1} = k \]
This gives:
\[ x = d + k, \quad y = 2d - k, \quad z = 3d + k \]

### Step 3: Substitute into the plane equation
Substituting these into the plane equation \( x - y + z = 1 \):
\[ (d + k) - (2d - k) + (3d + k) = 1 \]
Simplifying:
\[ d + k - 2d + k + 3d + k = 1 \implies 2d + 3k = 1 \implies k = \frac{1 - 2d}{3} \]

### Step 4: Find coordinates of foot \( N \)
Substituting \( k \) back into the parametric equations:
\[ x_1 = d + \frac{1 - 2d}{3} = \frac{3d + 1 - 2d}{3} = \frac{d + 1}{3} \]
\[ y_1 = 2d - \frac{1 - 2d}{3} = \frac{6d - 1 + 2d}{3} = \frac{8d - 1}{3} \]
\[ z_1 = 3d + \frac{1 - 2d}{3} = \frac{9d + 1 - 2d}{3} = \frac{7d + 1}{3} \]

### Step 5: Find the reflection point \( M \)
The reflection point \( M \) can be found using:
\[ M = (2x_1 - d, 2y_1 - 2d, 2z_1 - 3d) \]
Calculating each coordinate:
\[ M_x = 2 \cdot \frac{d + 1}{3} - d = \frac{2d + 2 - 3d}{3} = \frac{-d + 2}{3} \]
\[ M_y = 2 \cdot \frac{8d - 1}{3} - 2d = \frac{16d - 2 - 6d}{3} = \frac{10d - 2}{3} \]
\[ M_z = 2 \cdot \frac{7d + 1}{3} - 3d = \frac{14d + 2 - 9d}{3} = \frac{5d + 2}{3} \]

### Final Reflection Point
Thus, the reflection point \( M \) is:
\[ M\left(\frac{-d + 2}{3}, \frac{10d - 2}{3}, \frac{5d + 2}{3}\right) \]