To find the square of the length of the segment \( QS \) where \( Q \) is the image of point \( P(-1, 2, 3) \) in the plane \( x + y - z - 3 = 0 \) and \( S \) is a point on this plane with coordinates \( (3, 2, \beta) \), we will follow these steps:
### Step 1: Find the equation of the plane
The equation of the plane is given as:
\[
x + y - z - 3 = 0
\]
### Step 2: Identify the normal vector of the plane
The normal vector \( \vec{n} \) of the plane can be derived from the coefficients of \( x, y, z \) in the plane equation:
\[
\vec{n} = (1, 1, -1)
\]
### Step 3: Find the coordinates of the image point \( Q \)
To find the image of point \( P(-1, 2, 3) \) in the plane, we can use the formula for the image of a point in a plane:
\[
\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = -\frac{2(ax_1 + by_1 + cz_1 + d)}{a^2 + b^2 + c^2}
\]
where \( (x_1, y_1, z_1) \) are the coordinates of point \( P \) and \( (a, b, c) \) are the coefficients from the plane equation.
Here, \( a = 1, b = 1, c = -1, d = -3 \) and \( (x_1, y_1, z_1) = (-1, 2, 3) \).
Calculating \( ax_1 + by_1 + cz_1 + d \):
\[
= 1(-1) + 1(2) - 1(3) - 3 = -1 + 2 - 3 - 3 = -5
\]
Now, substituting into the formula:
\[
\frac{x + 1}{1} = \frac{y - 2}{1} = \frac{z - 3}{-1} = -\frac{2(-5)}{1^2 + 1^2 + (-1)^2} = -\frac{-10}{3} = \frac{10}{3}
\]
Setting \( k = \frac{10}{3} \):
\[
x + 1 = k \implies x = k - 1 = \frac{10}{3} - 1 = \frac{7}{3}
\]
\[
y - 2 = k \implies y = k + 2 = \frac{10}{3} + 2 = \frac{16}{3}
\]
\[
z - 3 = -k \implies z = 3 - k = 3 - \frac{10}{3} = -\frac{1}{3}
\]
Thus, the coordinates of point \( Q \) are:
\[
Q\left(\frac{7}{3}, \frac{16}{3}, -\frac{1}{3}\right)
\]
### Step 4: Find the value of \( \beta \) for point \( S \)
Point \( S \) has coordinates \( (3, 2, \beta) \) and must satisfy the plane equation:
\[
3 + 2 - \beta - 3 = 0 \implies 2 - \beta = 0 \implies \beta = 2
\]
Thus, the coordinates of point \( S \) are:
\[
S(3, 2, 2)
\]
### Step 5: Calculate the square of the distance \( QS \)
Using the distance formula:
\[
QS^2 = (x_Q - x_S)^2 + (y_Q - y_S)^2 + (z_Q - z_S)^2
\]
Substituting the coordinates of \( Q \) and \( S \):
\[
QS^2 = \left(\frac{7}{3} - 3\right)^2 + \left(\frac{16}{3} - 2\right)^2 + \left(-\frac{1}{3} - 2\right)^2
\]
Calculating each term:
1. \( \frac{7}{3} - 3 = \frac{7}{3} - \frac{9}{3} = -\frac{2}{3} \)
\[
\left(-\frac{2}{3}\right)^2 = \frac{4}{9}
\]
2. \( \frac{16}{3} - 2 = \frac{16}{3} - \frac{6}{3} = \frac{10}{3} \)
\[
\left(\frac{10}{3}\right)^2 = \frac{100}{9}
\]
3. \( -\frac{1}{3} - 2 = -\frac{1}{3} - \frac{6}{3} = -\frac{7}{3} \)
\[
\left(-\frac{7}{3}\right)^2 = \frac{49}{9}
\]
Combining these:
\[
QS^2 = \frac{4}{9} + \frac{100}{9} + \frac{49}{9} = \frac{4 + 100 + 49}{9} = \frac{153}{9} = 17
\]
### Final Answer
The square of the length of segment \( QS \) is:
\[
\boxed{17}
\]
