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

Explanation

To solve the problem, we need to determine the equations of two planes whose intersection is represented by the given plane equation \(2x - (1 + \lambda)y + 3\lambda z = 0\).

### Step-by-Step Solution:

1. **Identify the Given Plane Equation:**
The equation of the plane is given as:
\[ 2x - (1 + \lambda)y + 3\lambda z = 0 \]

2. **Rewrite the Plane Equation:**
We can rewrite the equation by factoring out \(\lambda\):
\[ 2x - y - \lambda y + 3\lambda z = 0 \]
This can be rearranged as:
\[ 2x - y + \lambda(-y + 3z) = 0 \]

3. **General Form of the Intersection of Two Planes:**
The general equation of a plane that passes through the intersection of two planes \(P_1\) and \(P_2\) can be expressed as:
\[ P_1 + \lambda P_2 = 0 \]
where \(P_1\) and \(P_2\) are the equations of the two planes.

4. **Identify the Coefficients:**
From our rewritten equation:
- The coefficient of \(x\) is \(2\).
- The coefficient of \(y\) is \(-1 - \lambda\).
- The coefficient of \(z\) is \(3\lambda\).

5. **Set Up the Equations for \(P_1\) and \(P_2\):**
We can assume:
\[ P_1: 2x - y = 0 \quad \text{(Equation 1)} \]
\[ P_2: -y + 3z = 0 \quad \text{(Equation 2)} \]

6. **Rearranging Equation 2:**
We can rearrange Equation 2:
\[ -y + 3z = 0 \implies y - 3z = 0 \]
This means:
\[ P_2: y - 3z = 0 \]

7. **Final Equations of the Planes:**
Thus, the two planes are:
\[ P_1: 2x - y = 0 \]
\[ P_2: y - 3z = 0 \]

8. **Conclusion:**
The correct option that matches the derived equations is the second option, which states:
\[ P_1: 2x - y = 0 \quad \text{and} \quad P_2: y - 3z = 0 \]