In R^3,consider the planes P_1: y=0 and P_2: | vedclass

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(C) The equation of any plane passing through the intersection of $P_1: y=0$ and $P_2: x+z-1=0$ is given by $(x+z-1) + \lambda y = 0$,which simplifies

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$(B, D)$

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(C) The equation of any plane passing through the intersection of $P_1: y=0$ and $P_2: x+z-1=0$ is given by $(x+z-1) + \lambda y = 0$,which simplifies to $x + \lambda y + z - 1 = 0$.The distance of the point $(0, 1, 0)$ from this plane is given as $1$:$\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$$-2\lambda = 1 \Rightarrow \lambda = -\frac{1}{2}$.Substituting $\lambda = -\frac{1}{2}$ into the plane equation:$x - \frac{1}{2}y + z - 1 = 0 \Rightarrow 2x - y + 2z - 2 = 0$.Now,the distance of the point $(\alpha, \beta, \gamma)$ from the plane $2x - y + 2z - 2 = 0$ is $2$:$\frac{|2\alpha - \beta + 2\gamma - 2|}{\sqrt{2^2 + (-1)^2 + 2^2}} = 2$$\frac{|2\alpha - \beta + 2\gamma - 2|}{3} = 2$$|2\alpha - \beta + 2\gamma - 2| = 6$.This gives two possibilities:$2\alpha - \beta + 2\gamma - 2 = 6 \Rightarrow 2\alpha - \beta + 2\gamma - 8 = 0$$2\alpha - \beta + 2\gamma - 2 = -6 \Rightarrow 2\alpha - \beta + 2\gamma + 4 = 0$.Thus,options $(B)$ and $(D)$ are correct.

$List-I$	$List-II$
$(P) \gamma$ equals	$(1) -\hat{i}-\hat{j}+\hat{k}$
$(Q)$ $A$ possible choice for $\hat{n}$ is	$(2) \sqrt{\frac{3}{2}}$
$(R) \vec{OR_1}$ equals	$(3) 1$
$(S)$ $A$ possible value of $\vec{OR_1} \cdot \hat{n}$ is	$(4) \frac{1}{\sqrt{6}} \hat{i}-\frac{2}{\sqrt{6}} \hat{j}+\frac{1}{\sqrt{6}} \hat{k}$
	$(5) \sqrt{\frac{2}{3}}$

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