Equation of Plane Passing Through Intersection of Two Planes

To find the equation of the plane $P_3$, we use the family of planes passing through the intersection of $P_1 : y = 0$ and $P_2 : x + z - 1 = 0$.

The general equation of such a plane $P_3$ (where $P_3$ is distinct from $P_1$) can be written as: $(x + z - 1) + \lambda y = 0$ Rearranging the terms, we get: $x + \lambda y + z - 1 = 0$

The perpendicular distance from the point $(0, 1, 0)$ to the plane $P_3$ is given to be $1$: $\frac{|0 + \lambda(1) + 0 - 1|}{\sqrt{1^2 + \lambda^2 + 1^2}} = 1$ $\frac{|\lambda - 1|}{\sqrt{\lambda^2 + 2}} = 1$

Squaring both sides: $(\lambda - 1)^2 = \lambda^2 + 2$ $\lambda^2 - 2\lambda + 1 = \lambda^2 + 2$ $-2\lambda = 1 \implies \lambda = -\frac{1}{2}$

Substitute $\lambda = -\frac{1}{2}$ back into the equation of $P_3$: $x - \frac{1}{2}y + z - 1 = 0$ Multiplying the entire equation by $2$, we obtain the equation of plane $P_3$: $2x - y + 2z - 2 = 0$

Now, it is given that the perpendicular distance from the point $(\alpha, \beta, \gamma)$ to $P_3$ is $2$: $\frac{|2\alpha - \beta + 2\gamma - 2|}{\sqrt{2^2 + (-1)^2 + 2^2}} = 2$ $\frac{|2\alpha - \beta + 2\gamma - 2|}{\sqrt{4 + 1 + 4}} = 2$ $\frac{|2\alpha - \beta + 2\gamma - 2|}{3} = 2$ $|2\alpha - \beta + 2\gamma - 2| = 6$

This leads to two possibilities:

1. $2\alpha - \beta + 2\gamma - 2 = 6 \implies 2\alpha - \beta + 2\gamma - 8 = 0$
2. $2\alpha - \beta + 2\gamma - 2 = -6 \implies 2\alpha - \beta + 2\gamma + 4 = 0$

Thus, the relations that can be true are given by options (B) and (D).