In `R^(3)`, consider the planes `P_(1):y=0` and `P_(2),x+z=1.` Let `P_(3)` be a plane, different from `P_(1)` and `P_(2)` which passes through the intersection of `P_(1)` and `P_(2)`, If the distance of the point (0,1,0) from `P_(3)` is 1 and the distance of a point `(alpha,beta,gamma)` from `P_(3)` is 2, then which of the following relation(s) is/are true?

Options

A. `2alpha+beta+2gamma+2=0`
B. `2alpha+beta+2gamma+4=0`
C. `2alpha+beta+2gamma-10=0`
D. `2alpha+beta+2gamma-8=0`

Explanation

To solve the problem step by step, we will first derive the equation of the plane \( P_3 \) and then use the distance formula to find the required relations.

### Step 1: Find the equation of plane \( P_3 \)

Given the planes:
- \( P_1: y = 0 \) (which can be written as \( 0x + 1y + 0z = 0 \))
- \( P_2: x + z = 1 \) (which can be written as \( 1x + 0y + 1z = 1 \))

Since \( P_3 \) passes through the intersection of \( P_1 \) and \( P_2 \), we can express \( P_3 \) as a linear combination of \( P_1 \) and \( P_2 \):
\[ P_3: x + \lambda y + z - 1 = 0 \]
where \( \lambda \) is a parameter.

### Step 2: Use the distance formula from point \( (0, 1, 0) \) to plane \( P_3 \)

The distance \( d \) from a point \( (x_0, y_0, z_0) \) to a 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 the point \( (0, 1, 0) \) and the plane \( P_3: x + \lambda y + z - 1 = 0 \), we have:
- \( A = 1 \)
- \( B = \lambda \)
- \( C = 1 \)
- \( D = -1 \)

Substituting into the distance formula:
\[ d = \frac{|1(0) + \lambda(1) + 1(0) - 1|}{\sqrt{1^2 + \lambda^2 + 1^2}} = \frac{|\lambda - 1|}{\sqrt{2 + \lambda^2}} \]

We know this distance is equal to 1:
\[ \frac{|\lambda - 1|}{\sqrt{2 + \lambda^2}} = 1 \]

### Step 3: Solve the equation for \( \lambda \)

Cross-multiplying gives:
\[ |\lambda - 1| = \sqrt{2 + \lambda^2} \]

Squaring both sides:
\[ (\lambda - 1)^2 = 2 + \lambda^2 \]
Expanding the left side:
\[ \lambda^2 - 2\lambda + 1 = 2 + \lambda^2 \]
Cancelling \( \lambda^2 \) from both sides:
\[ -2\lambda + 1 = 2 \]
Solving for \( \lambda \):
\[ -2\lambda = 1 \implies \lambda = -\frac{1}{2} \]

### Step 4: Find the distance from point \( (\alpha, \beta, \gamma) \) to plane \( P_3 \)

Now we need to find the distance from the point \( (\alpha, \beta, \gamma) \) to the plane \( P_3 \):
\[ d = \frac{|1\alpha + (-\frac{1}{2})\beta + 1\gamma - 1|}{\sqrt{1^2 + (-\frac{1}{2})^2 + 1^2}} = 2 \]

Calculating the denominator:
\[ \sqrt{1 + \frac{1}{4} + 1} = \sqrt{\frac{9}{4}} = \frac{3}{2} \]

Thus, we have:
\[ \frac{|\alpha - \frac{1}{2}\beta + \gamma - 1|}{\frac{3}{2}} = 2 \]

Cross-multiplying gives:
\[ |\alpha - \frac{1}{2}\beta + \gamma - 1| = 3 \]

### Step 5: Solve for the relations

This leads to two cases:
1. \( \alpha - \frac{1}{2}\beta + \gamma - 1 = 3 \)
2. \( \alpha - \frac{1}{2}\beta + \gamma - 1 = -3 \)

From the first case:
\[ \alpha - \frac{1}{2}\beta + \gamma = 4 \]

From the second case:
\[ \alpha - \frac{1}{2}\beta + \gamma = -2 \]

### Final Relations

Thus, we have two relations:
1. \( 2\alpha - \beta + 2\gamma - 8 = 0 \)
2. \( 2\alpha - \beta + 2\gamma + 4 = 0 \)