To solve the problem step by step, we need to find the equation of the plane containing the incident ray and the reflected ray, and then calculate the distance of that plane from the origin (0, 0, 0). Finally, we will compute the value of \( 13\lambda^2 \).
### Step 1: Identify the points and the plane
We have:
- Point A: \( A(1, 2, 3) \)
- Point C: \( C(3, 5, 9) \)
- Plane equation: \( x + y + z = 12 \)
### Step 2: Find the normal vector of the plane
The normal vector \( \vec{n} \) of the plane \( x + y + z = 12 \) is given by the coefficients of \( x, y, z \):
\[
\vec{n} = (1, 1, 1)
\]
### Step 3: Find the direction vector of the incident ray
The direction vector of the ray from point A to point C is:
\[
\vec{AC} = C - A = (3 - 1, 5 - 2, 9 - 3) = (2, 3, 6)
\]
### Step 4: Find the point of intersection B on the plane
Let the point of intersection B be \( (x, y, z) \). Since B lies on the plane, it satisfies the equation:
\[
x + y + z = 12
\]
We can express \( z \) in terms of \( x \) and \( y \):
\[
z = 12 - x - y
\]
### Step 5: Set up the equation of the plane containing the rays
The plane containing the incident ray (from A to B) and the reflected ray (from B to C) can be determined using the points A, B, and C. The equation of the plane can be derived from the determinant:
\[
\begin{vmatrix}
x - 1 & y - 2 & z - 3 \\
2 & 3 & 6 \\
1 & 1 & 1
\end{vmatrix} = 0
\]
### Step 6: Expand the determinant
Expanding the determinant gives:
\[
(x - 1) \begin{vmatrix} 3 & 6 \\ 1 & 1 \end{vmatrix} - (y - 2) \begin{vmatrix} 2 & 6 \\ 1 & 1 \end{vmatrix} + (z - 3) \begin{vmatrix} 2 & 3 \\ 1 & 1 \end{vmatrix} = 0
\]
Calculating the minors:
\[
= (x - 1)(3 - 6) - (y - 2)(2 - 6) + (z - 3)(2 - 3) = 0
\]
\[
= -3(x - 1) + 4(y - 2) - (z - 3) = 0
\]
\[
= -3x + 3 + 4y - 8 - z + 3 = 0
\]
\[
\Rightarrow -3x + 4y - z - 2 = 0
\]
### Step 7: Rearranging the equation
Rearranging gives us:
\[
3x - 4y + z + 2 = 0
\]
### Step 8: Calculate the distance from the origin to the plane
The distance \( d \) from the origin \( (0, 0, 0) \) to the plane \( Ax + By + Cz + D = 0 \) is given by:
\[
d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}
\]
Here, \( A = 3, B = -4, C = 1, D = 2 \):
\[
d = \frac{|2|}{\sqrt{3^2 + (-4)^2 + 1^2}} = \frac{2}{\sqrt{9 + 16 + 1}} = \frac{2}{\sqrt{26}}
\]
### Step 9: Set \( \lambda \)
Let \( \lambda = \frac{2}{\sqrt{26}} \).
### Step 10: Calculate \( 13\lambda^2 \)
Now we compute:
\[
\lambda^2 = \left(\frac{2}{\sqrt{26}}\right)^2 = \frac{4}{26} = \frac{2}{13}
\]
Thus,
\[
13\lambda^2 = 13 \times \frac{2}{13} = 2
\]
### Final Answer
The value of \( 13\lambda^2 \) is \( \boxed{2} \).
