A ray of light through the point `A(1,2,3)` strikes the plane `x+y+z=12` at a point B and on reflection passes through the point `C(3, 5, 9)`. If the equation of a plane containing the incident ray and the reflected ray is P = 0 has the distance of P = 0 from (0, 0, 0) is `lambda` units, then the value of `13lambda^(2)` is equal to

Options

A. 1
B. 2
C. 4
D. 6

Answer

To solve the problem step by step, we will follow the reasoning laid out in the video transcript while ensuring clarity in each step.

### Step 1: Identify Points and the Plane
We have three points:
- Point A: \( A(1, 2, 3) \) (the point from which the ray of light originates)
- Point B: \( B(x_B, y_B, z_B) \) (the point where the ray strikes the plane)
- Point C: \( C(3, 5, 9) \) (the point through which the reflected ray passes)

The equation of the plane is given as:
\[ x + y + z = 12 \]

### 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: Determine the Direction Vectors
The direction vector of the incident ray \( \vec{d_1} \) from point A to point B can be expressed as:
\[ \vec{d_1} = (x_B - 1, y_B - 2, z_B - 3) \]

The direction vector of the reflected ray \( \vec{d_2} \) from point B to point C is:
\[ \vec{d_2} = (3 - x_B, 5 - y_B, 9 - z_B) \]

### Step 4: Use the Reflection Law
For reflection, the angle of incidence equals the angle of reflection. This implies that the direction vectors \( \vec{d_1} \) and \( \vec{d_2} \) must satisfy the condition:
\[ \vec{d_1} + \vec{d_2} = k \vec{n} \]
for some scalar \( k \).

### Step 5: Set Up the Equation
Substituting the direction vectors:
\[ (x_B - 1 + 3 - x_B, y_B - 2 + 5 - y_B, z_B - 3 + 9 - z_B) = k(1, 1, 1) \]
This simplifies to:
\[ (2, 3, 6) = k(1, 1, 1) \]
From this, we find \( k = 2 \).

### Step 6: Find the Equation of the Plane Containing the Rays
The plane containing the rays can be expressed in the form:
\[ \vec{n} \cdot ( (x, y, z) - (1, 2, 3) ) = 0 \]
Substituting the normal vector:
\[ 1(x - 1) + 1(y - 2) + 1(z - 3) = 0 \]
This simplifies to:
\[ x + y + z - 6 = 0 \]
or
\[ x + y + z = 6 \]

### Step 7: Find the Distance from the Origin to the Plane
The distance \( d \) from the origin \( (0, 0, 0) \) to the plane \( x + y + z = 6 \) is given by the formula:
\[ d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}} \]
where \( D \) is the constant term (6), and \( A, B, C \) are the coefficients of \( x, y, z \) respectively (which are all 1).

Calculating:
\[ d = \frac{|6|}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{6}{\sqrt{3}} = 2\sqrt{3} \]

### Step 8: Relate Distance to Lambda
We have defined the distance from the origin as \( \lambda \):
\[ \lambda = 2\sqrt{3} \]

### Step 9: Calculate \( 13\lambda^2 \)
Now we need to find \( 13\lambda^2 \):
\[ \lambda^2 = (2\sqrt{3})^2 = 4 \times 3 = 12 \]
Thus,
\[ 13\lambda^2 = 13 \times 12 = 156 \]

### Final Answer
The value of \( 13\lambda^2 \) is:
\[ \boxed{156} \]