To solve the problem, we need to find the equation of the plane \( P_3 \) that passes through the intersection of the planes \( P_1: y = 0 \) and \( P_2: x + z = 1 \). We will then use the distance formula to derive the necessary conditions for the points given in the problem.
### Step 1: Identify the equations of the planes
The equations of the planes are given as:
- Plane \( P_1: y = 0 \)
- Plane \( P_2: x + z = 1 \)
### Step 2: Find the equation of plane \( P_3 \)
Since plane \( P_3 \) passes through the intersection of \( P_1 \) and \( P_2 \), we can express the equation of \( P_3 \) in the form:
\[
P_3: x + z - 1 + \lambda y = 0
\]
where \( \lambda \) is a parameter.
### Step 3: Calculate the distance from the point \( (0, 1, 0) \) to plane \( P_3 \)
The distance \( d \) from a point \( (x_0, y_0, z_0) \) to the plane \( Ax + By + Cz + D = 0 \) is given by:
\[
d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}
\]
For our plane \( P_3: x + z - 1 + \lambda y = 0 \), we have:
- \( A = 1 \), \( B = \lambda \), \( C = 1 \), \( D = -1 \)
Substituting the point \( (0, 1, 0) \):
\[
d = \frac{|1 \cdot 0 + \lambda \cdot 1 + 1 \cdot 0 - 1|}{\sqrt{1^2 + \lambda^2 + 1^2}} = \frac{|\lambda - 1|}{\sqrt{2 + \lambda^2}}
\]
According to the problem, this distance is equal to 1:
\[
\frac{|\lambda - 1|}{\sqrt{2 + \lambda^2}} = 1
\]
### Step 4: Solve the distance equation
Squaring both sides, we get:
\[
|\lambda - 1| = \sqrt{2 + \lambda^2}
\]
This leads to two cases:
1. \( \lambda - 1 = \sqrt{2 + \lambda^2} \)
2. \( \lambda - 1 = -\sqrt{2 + \lambda^2} \)
#### Case 1: \( \lambda - 1 = \sqrt{2 + \lambda^2} \)
Squaring both sides:
\[
(\lambda - 1)^2 = 2 + \lambda^2
\]
Expanding and simplifying:
\[
\lambda^2 - 2\lambda + 1 = 2 + \lambda^2 \implies -2\lambda + 1 - 2 = 0 \implies -2\lambda - 1 = 0 \implies \lambda = -\frac{1}{2}
\]
#### Case 2: \( \lambda - 1 = -\sqrt{2 + \lambda^2} \)
Squaring both sides:
\[
(\lambda - 1)^2 = 2 + \lambda^2
\]
This leads to the same equation as above, confirming \( \lambda = -\frac{1}{2} \).
### Step 5: Substitute \( \lambda \) back to find the equation of \( P_3 \)
Substituting \( \lambda = -\frac{1}{2} \) into the equation of \( P_3 \):
\[
P_3: x + z - 1 - \frac{1}{2}y = 0 \implies 2x - y + 2z - 2 = 0
\]
### Step 6: Calculate the distance from point \( (\alpha, \beta, \gamma) \) to plane \( P_3 \)
Using the distance formula again for point \( (\alpha, \beta, \gamma) \):
\[
d = \frac{|2\alpha - \beta + 2\gamma - 2|}{\sqrt{2^2 + (-1)^2 + 2^2}} = \frac{|2\alpha - \beta + 2\gamma - 2|}{3}
\]
Given that this distance is equal to 2:
\[
\frac{|2\alpha - \beta + 2\gamma - 2|}{3} = 2 \implies |2\alpha - \beta + 2\gamma - 2| = 6
\]
This leads to two equations:
1. \( 2\alpha - \beta + 2\gamma - 2 = 6 \)
2. \( 2\alpha - \beta + 2\gamma - 2 = -6 \)
### Step 7: Solve the equations
From \( 2\alpha - \beta + 2\gamma - 2 = 6 \):
\[
2\alpha - \beta + 2\gamma - 8 = 0 \quad \text{(Equation 1)}
\]
From \( 2\alpha - \beta + 2\gamma - 2 = -6 \):
\[
2\alpha - \beta + 2\gamma + 4 = 0 \quad \text{(Equation 2)}
\]
### Conclusion: Identify the correct options
The equations derived are:
1. \( 2\alpha - \beta + 2\gamma - 8 = 0 \) (Option d)
2. \( 2\alpha - \beta + 2\gamma + 4 = 0 \) (Option b)
Thus, the correct options are (b) and (d).
