Step-by-step Solution:

1. Find the general equation of plane P₃

   The plane $P_3$ passes through the intersection of planes $P_1: y = 0$ and $P_2: x + z = 1$. The equation of any such plane can be written as a linear combination of the equations of $P_1$ and $P_2$.

   The equation of the family of planes is given by $P_1 + \lambda P_2 = 0$, where $\lambda$ is a scalar parameter.

   Substituting the equations of $P_1$ and $P_2$: $y + \lambda (x + z - 1) = 0$ This can be rearranged to the standard form $Ax+By+Cz+D=0$: $\lambda x + y + \lambda z - \lambda = 0$ The problem states that $P_3$ is different from $P_1$ and $P_2$. If $\lambda = 0$, the equation becomes $y=0$, which is $P_1$. So, we must have $\lambda \neq 0$.

2. Use the first condition to find the value of λ

   The distance of the point $(0, 1, 0)$ from the plane $P_3$ is given as $1$. The formula for the distance of a point $(x_0, y_0, z_0)$ from a plane $Ax+By+Cz+D=0$ is: $d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$ For plane $P_3: \lambda x + y + \lambda z - \lambda = 0$ and point $(0, 1, 0)$, we have: $1 = \frac{|\lambda(0) + 1(1) + \lambda(0) - \lambda|}{\sqrt{\lambda^2 + 1^2 + \lambda^2}}$ $1 = \frac{|1 - \lambda|}{\sqrt{2\lambda^2 + 1}}$ Squaring both sides to remove the square root and the absolute value: $1^2 = \frac{(1 - \lambda)^2}{2\lambda^2 + 1}$ $2\lambda^2 + 1 = (1 - \lambda)^2$ $2\lambda^2 + 1 = 1 - 2\lambda + \lambda^2$ $2\lambda^2 - \lambda^2 + 2\lambda + 1 - 1 = 0$ $\lambda^2 + 2\lambda = 0$ $\lambda(\lambda + 2) = 0$ This yields two possible values for $\lambda$: $\lambda = 0$ or $\lambda = -2$. As established in Step 1, $\lambda \neq 0$ because $P_3$ is different from $P_1$. Therefore, we must have $\lambda = -2$.

3. Determine the specific equation of plane P₃

   Substitute $\lambda = -2$ into the general equation of $P_3$: $-2x + y - 2z - (-2) = 0$ $-2x + y - 2z + 2 = 0$ Multiplying the entire equation by $-1$ gives a more conventional form with a positive leading coefficient: $2x - y + 2z - 2 = 0$

4. Use the second condition to find the relations for (α, β, γ)

   The distance of a point $(\alpha, \beta, \gamma)$ from the plane $P_3: 2x - y + 2z - 2 = 0$ is given as $2$. Using the distance formula again: $2 = \frac{|2(\alpha) - 1(\beta) + 2(\gamma) - 2|}{\sqrt{2^2 + (-1)^2 + 2^2}}$ $2 = \frac{|2\alpha - \beta + 2\gamma - 2|}{\sqrt{4 + 1 + 4}}$ $2 = \frac{|2\alpha - \beta + 2\gamma - 2|}{\sqrt{9}}$ $2 = \frac{|2\alpha - \beta + 2\gamma - 2|}{3}$ $6 = |2\alpha - \beta + 2\gamma - 2|$

5. Derive the two possible relations

   The absolute value equation leads to two possible linear equations:

   • Case 1: $2\alpha - \beta + 2\gamma - 2 = 6$ $2\alpha - \beta + 2\gamma - 8 = 0$ This matches option (D).

   • Case 2: $2\alpha - \beta + 2\gamma - 2 = -6$ $2\alpha - \beta + 2\gamma + 4 = 0$ This matches option (B).

Conclusion

Both relations derived from the given conditions are listed in the options. Therefore, options (B) and (D) are the correct answers.