To find the image of the point \( P(1, 2, -3) \) in the line given by the equation
\[
\frac{x + 1}{2} = \frac{y - 3}{-2} = \frac{z}{-1},
\]
we will follow these steps:
### Step 1: Parametrize the line
We can express the coordinates of points on the line in terms of a parameter \( \lambda \):
\[
x = 2\lambda - 1,
\]
\[
y = -2\lambda + 3,
\]
\[
z = -\lambda.
\]
### Step 2: Find the coordinates of the midpoint \( O \)
Let \( O(a, b, c) \) be the midpoint of the segment \( PQ \), where \( P(1, 2, -3) \) is the original point and \( Q(a, b, c) \) is its image. The coordinates of the midpoint \( O \) can be expressed as:
\[
O = \left( \frac{1 + a}{2}, \frac{2 + b}{2}, \frac{-3 + c}{2} \right).
\]
### Step 3: Set the midpoint equal to the coordinates from the line
Since \( O \) lies on the line, we can equate the coordinates of \( O \) from the line with those from the midpoint:
\[
\frac{1 + a}{2} = 2\lambda - 1,
\]
\[
\frac{2 + b}{2} = -2\lambda + 3,
\]
\[
\frac{-3 + c}{2} = -\lambda.
\]
### Step 4: Solve for \( a, b, c \) in terms of \( \lambda \)
From the first equation:
\[
1 + a = 4\lambda - 2 \implies a = 4\lambda - 3,
\]
From the second equation:
\[
2 + b = -4\lambda + 6 \implies b = -4\lambda + 4,
\]
From the third equation:
\[
-3 + c = -2\lambda \implies c = -2\lambda + 3.
\]
### Step 5: Find the direction ratios of the line
The direction ratios of the line are \( (2, -2, -1) \). The vector \( OP \) (from \( O \) to \( P \)) can be expressed as:
\[
OP = (1 - (2\lambda - 1), 2 - (-2\lambda + 3), -3 - (-\lambda)).
\]
This simplifies to:
\[
OP = (2 - 2\lambda, -2 + 2\lambda, -3 + \lambda).
\]
### Step 6: Use the perpendicularity condition
Since \( OP \) is perpendicular to the line, we apply the dot product condition:
\[
(2, -2, -1) \cdot (2 - 2\lambda, -2 + 2\lambda, -3 + \lambda) = 0.
\]
Calculating the dot product:
\[
2(2 - 2\lambda) + (-2)(-2 + 2\lambda) + (-1)(-3 + \lambda) = 0.
\]
Expanding this gives:
\[
4 - 4\lambda + 4 - 4\lambda + 3 - \lambda = 0 \implies 11 - 9\lambda = 0.
\]
### Step 7: Solve for \( \lambda \)
From \( 11 - 9\lambda = 0 \):
\[
\lambda = \frac{11}{9}.
\]
### Step 8: Substitute \( \lambda \) back to find \( a, b, c \)
Substituting \( \lambda \) back into the equations for \( a, b, c \):
\[
a = 4\left(\frac{11}{9}\right) - 3 = \frac{44}{9} - \frac{27}{9} = \frac{17}{9},
\]
\[
b = -4\left(\frac{11}{9}\right) + 4 = -\frac{44}{9} + \frac{36}{9} = -\frac{8}{9},
\]
\[
c = -2\left(\frac{11}{9}\right) + 3 = -\frac{22}{9} + \frac{27}{9} = \frac{5}{9}.
\]
### Step 9: Calculate \( a + b + c \)
Now, we can find \( a + b + c \):
\[
a + b + c = \frac{17}{9} - \frac{8}{9} + \frac{5}{9} = \frac{17 - 8 + 5}{9} = \frac{14}{9}.
\]
### Final Answer
Thus, the value of \( a + b + c \) is:
\[
\boxed{\frac{14}{9}}.
\]
