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

Options

A. (1,1,7)
B. (0,1,7)
C. (1,0,7)
D. none of these

Explanation

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

### Step 1: Parametrize the line
The line can be expressed in parametric form. Let \( \lambda \) be the parameter. Then, the coordinates of any point \( Q \) on the line can be written as:
\[ Q(\lambda) = (1\lambda, 2\lambda + 1, 3\lambda + 2) \]
This gives us:
- \( x = \lambda \)
- \( y = 2\lambda + 1 \)
- \( z = 3\lambda + 2 \)

### Step 2: Find the foot of the perpendicular
Let \( Q \) be the foot of the perpendicular from point \( P(1, 6, 3) \) to the line. The vector \( \overrightarrow{PQ} \) can be expressed as:
\[ \overrightarrow{PQ} = (x - 1, y - 6, z - 3) \]
Substituting the parametric coordinates of \( Q \):
\[ \overrightarrow{PQ} = (\lambda - 1, (2\lambda + 1) - 6, (3\lambda + 2) - 3) \]
This simplifies to:
\[ \overrightarrow{PQ} = (\lambda - 1, 2\lambda - 5, 3\lambda - 1) \]

### Step 3: Direction vector of the line
The direction vector of the line is \( \overrightarrow{d} = (1, 2, 3) \).

### Step 4: Set up the dot product for perpendicularity
Since \( \overrightarrow{PQ} \) and \( \overrightarrow{d} \) are perpendicular, their dot product must equal zero:
\[ (\lambda - 1) \cdot 1 + (2\lambda - 5) \cdot 2 + (3\lambda - 1) \cdot 3 = 0 \]
Expanding this gives:
\[ \lambda - 1 + 4\lambda - 10 + 9\lambda - 3 = 0 \]
Combining like terms results in:
\[ 14\lambda - 14 = 0 \]
Thus, solving for \( \lambda \):
\[ \lambda = 1 \]

### Step 5: Find the coordinates of the foot of the perpendicular \( Q \)
Substituting \( \lambda = 1 \) back into the parametric equations:
\[ Q(1) = (1, 2(1) + 1, 3(1) + 2) = (1, 3, 5) \]

### Step 6: Use the midpoint formula
Let \( R(x, y, z) \) be the image of point \( P \) across point \( Q \). Since \( Q \) is the midpoint of \( P \) and \( R \), we can use the midpoint formula:
\[ Q = \left( \frac{x + 1}{2}, \frac{y + 6}{2}, \frac{z + 3}{2} \right) \]
Setting this equal to \( Q(1, 3, 5) \):
1. For \( x \):
\[ \frac{x + 1}{2} = 1 \implies x + 1 = 2 \implies x = 1 \]
2. For \( y \):
\[ \frac{y + 6}{2} = 3 \implies y + 6 = 6 \implies y = 0 \]
3. For \( z \):
\[ \frac{z + 3}{2} = 5 \implies z + 3 = 10 \implies z = 7 \]

### Final Answer
Thus, the image of point \( P(1, 6, 3) \) in the line is:
\[ R(1, 0, 7) \]