The plane `2x-(1+lambda)y+3lambdaz=0` passes through the intersection of the plane

Options

A. `2x-y=0` and `y-3z=0`
B. `2x+3z=0` and `y=0`
C. `2x-y+3z=0` and `y-3z=0`
D. none of these

Explanation

To solve the problem, we need to analyze the given plane equation and determine the planes that intersect to form it. Let's break down the solution step by step.

### Step 1: Write down the given plane equation
The equation of the plane is given as:
\[ 2x - (1 + \lambda)y + 3\lambda z = 0 \]

### Step 2: Rearrange the equation
We can rearrange the equation to isolate the terms involving \(\lambda\):
\[ 2x - y - \lambda y + 3\lambda z = 0 \]

### Step 3: Factor out \(\lambda\)
Next, we can group the terms involving \(\lambda\):
\[ 2x - y + \lambda(3z - y) = 0 \]

### Step 4: Identify the family of planes
This equation can be interpreted as a family of planes of the form:
\[ p_1 + \lambda p_2 = 0 \]
where:
- \( p_1 = 2x - y = 0 \) (Plane 1)
- \( p_2 = 3z - y = 0 \) (Plane 2)

### Step 5: Write the equations of the intersecting planes
From the identified planes, we can write the equations:
1. Plane 1: \( 2x - y = 0 \)
2. Plane 2: \( 3z - y = 0 \)

### Step 6: Conclusion
The planes that intersect to form the given plane are:
- \( 2x - y = 0 \)
- \( y - 3z = 0 \)

Thus, the final answer is:
- The two planes are \( 2x - y = 0 \) and \( y - 3z = 0 \).