Find the image of the point `(1, 2, 3)` in the line `(x-6)/(3)=(y-7)/(2)=(z-7)/(-2)`.

Answer

To find the image of the point \( A(1, 2, 3) \) in the line given by the equations \( \frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2} \), we can follow these steps:

### Step 1: Parameterize the Line
The line can be expressed in parametric form. Let \( \lambda \) be the parameter. The equations can be rewritten as:
\[ x = 3\lambda + 6, \quad y = 2\lambda + 7, \quad z = -2\lambda + 7 \]

### Step 2: Identify Point A
Let \( A(1, 2, 3) \) be the point from which we want to find the image.

### Step 3: Find Direction Ratios
The direction ratios of the line are \( (3, 2, -2) \). The vector \( \overrightarrow{AO} \) from point \( A \) to a point \( O \) on the line can be expressed as:
\[ \overrightarrow{AO} = (3\lambda + 6 - 1, 2\lambda + 7 - 2, -2\lambda + 7 - 3) = (3\lambda + 5, 2\lambda + 5, -2\lambda + 4) \]

### Step 4: Set Up the Perpendicular Condition
For the point \( O \) on the line to be the foot of the perpendicular from point \( A \), the dot product of \( \overrightarrow{AO} \) and the direction ratios of the line must be zero:
\[ (3\lambda + 5, 2\lambda + 5, -2\lambda + 4) \cdot (3, 2, -2) = 0 \]

### Step 5: Calculate the Dot Product
Calculating the dot product:
\[ 3(3\lambda + 5) + 2(2\lambda + 5) - 2(-2\lambda + 4) = 0 \]
Expanding this gives:
\[ 9\lambda + 15 + 4\lambda + 10 + 4\lambda - 8 = 0 \]
Combining like terms:
\[ (9\lambda + 4\lambda + 4\lambda) + (15 + 10 - 8) = 0 \implies 17\lambda + 17 = 0 \]

### Step 6: Solve for \( \lambda \)
Solving for \( \lambda \):
\[ 17\lambda = -17 \implies \lambda = -1 \]

### Step 7: Find Point O
Substituting \( \lambda = -1 \) back into the parametric equations to find point \( O \):
\[ x = 3(-1) + 6 = 3, \quad y = 2(-1) + 7 = 5, \quad z = -2(-1) + 7 = 9 \]
Thus, \( O(3, 5, 9) \).

### Step 8: Find Image Point B
Since \( O \) is the midpoint of \( A \) and its image \( B \), we can use the midpoint formula:
\[ O = \left( \frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}, \frac{z_A + z_B}{2} \right) \]
Substituting the known values:
\[ (3, 5, 9) = \left( \frac{1 + x_B}{2}, \frac{2 + y_B}{2}, \frac{3 + z_B}{2} \right) \]
This gives us three equations:
1. \( \frac{1 + x_B}{2} = 3 \)
2. \( \frac{2 + y_B}{2} = 5 \)
3. \( \frac{3 + z_B}{2} = 9 \)

### Step 9: Solve for \( B \)
Solving these equations:
1. \( 1 + x_B = 6 \implies x_B = 5 \)
2. \( 2 + y_B = 10 \implies y_B = 8 \)
3. \( 3 + z_B = 18 \implies z_B = 15 \)

Thus, the image point \( B \) is \( (5, 8, 15) \).

### Final Answer
The image of the point \( (1, 2, 3) \) in the line is \( (5, 8, 15) \).