To find the image of the point \( P(1, 6, 3) \) with respect to the line given by the equations:
\[
\frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z + 2}{1}
\]
we will follow these steps:
### Step 1: Determine the direction ratios of the line
From the line equation, we can identify the direction ratios as:
- \( a = -3 \)
- \( b = 2 \)
- \( c = 1 \)
### Step 2: Parametrize the line
Let \( \lambda \) be the parameter. The coordinates of any point \( O \) on the line can be expressed as:
\[
O = (-1 - 3\lambda, 3 + 2\lambda, -2 + \lambda)
\]
### Step 3: Find the vector \( OP \)
The vector \( OP \) from point \( O \) to point \( P(1, 6, 3) \) is given by:
\[
OP = P - O = \left(1 - (-1 - 3\lambda), 6 - (3 + 2\lambda), 3 - (-2 + \lambda)\right)
\]
This simplifies to:
\[
OP = (1 + 1 + 3\lambda, 6 - 3 - 2\lambda, 3 + 2 - \lambda) = (2 + 3\lambda, 3 - 2\lambda, 5 - \lambda)
\]
### Step 4: Find the normal vector to the line
The direction vector of the line is \( \vec{d} = (-3, 2, 1) \).
### Step 5: Set up the dot product condition
Since \( OP \) is perpendicular to the line, we can set up the dot product:
\[
OP \cdot \vec{d} = 0
\]
This gives:
\[
(2 + 3\lambda)(-3) + (3 - 2\lambda)(2) + (5 - \lambda)(1) = 0
\]
Expanding this:
\[
-6 - 9\lambda + 6 - 4\lambda + 5 - \lambda = 0
\]
Combining like terms:
\[
-14\lambda + 5 = 0
\]
Thus, solving for \( \lambda \):
\[
\lambda = \frac{5}{14}
\]
### Step 6: Find the coordinates of point \( O \)
Substituting \( \lambda = \frac{5}{14} \) back into the parametric equations for \( O \):
\[
O = \left(-1 - 3\left(\frac{5}{14}\right), 3 + 2\left(\frac{5}{14}\right), -2 + \left(\frac{5}{14}\right)\right)
\]
Calculating each coordinate:
- \( x = -1 - \frac{15}{14} = -\frac{29}{14} \)
- \( y = 3 + \frac{10}{14} = \frac{52}{14} \)
- \( z = -2 + \frac{5}{14} = -\frac{23}{14} \)
Thus, the coordinates of point \( O \) are:
\[
O\left(-\frac{29}{14}, \frac{52}{14}, -\frac{23}{14}\right)
\]
### Step 7: Use the midpoint formula to find the image point \( I \)
Let the image point be \( I(\alpha, \beta, \gamma) \). Since \( O \) is the midpoint of \( P \) and \( I \):
\[
\frac{1 + \alpha}{2} = -\frac{29}{14}, \quad \frac{6 + \beta}{2} = \frac{52}{14}, \quad \frac{3 + \gamma}{2} = -\frac{23}{14}
\]
Solving these equations:
1. For \( x \):
\[
1 + \alpha = -\frac{58}{14} \implies \alpha = -\frac{58}{14} - 1 = -\frac{58 + 14}{14} = -\frac{72}{14} = -\frac{36}{7}
\]
2. For \( y \):
\[
6 + \beta = \frac{104}{14} \implies \beta = \frac{104}{14} - 6 = \frac{104 - 84}{14} = \frac{20}{14} = \frac{10}{7}
\]
3. For \( z \):
\[
3 + \gamma = -\frac{46}{14} \implies \gamma = -\frac{46}{14} - 3 = -\frac{46 + 42}{14} = -\frac{88}{14} = -\frac{44}{7}
\]
### Final Result
Thus, the image of the point \( P(1, 6, 3) \) with respect to the line is:
\[
I\left(-\frac{36}{7}, \frac{10}{7}, -\frac{44}{7}\right)
\]
