Let ell_1 and ell_2 be the lines vec{r}_1=lam | vedclass

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(B) Given lines are $L_1: \vec{r}_1 = \lambda(\hat{i}+\hat{j}+\hat{k})$ and $L_2: \vec{r}_2 = (\hat{j}-\hat{k}) + \mu(\hat{i}+\hat{k})$.Any plane $H$

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$(P) \rightarrow (5), (Q) \rightarrow (4), (R) \rightarrow (3), (S) \rightarrow (1)$

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(B) Given lines are $L_1: \vec{r}_1 = \lambda(\hat{i}+\hat{j}+\hat{k})$ and $L_2: \vec{r}_2 = (\hat{j}-\hat{k}) + \mu(\hat{i}+\hat{k})$.Any plane $H$ containing $L_1$ passes through the origin $(0,0,0)$ and has a normal vector $\vec{n}$ perpendicular to the direction vector $(1,1,1)$ of $L_1$. Let the plane be $ax+by+cz=0$,where $a+b+c=0$.The distance $d(H)$ from $L_2$ to $H$ is non-zero only if $L_2$ is parallel to $H$. If $L_2$ is not parallel to $H$,the distance is $0$. To maximize $d(H)$,$L_2$ must be parallel to $H$. Thus,the normal $\vec{n} = (a,b,c)$ must be perpendicular to the direction vector of $L_2$,which is $(1,0,1)$.So,$a+c=0$. Since $a+b+c=0$ and $a+c=0$,we get $b=0$. Let $a=1$,then $c=-1$. The plane $H_0$ is $x-z=0$.$(P)$ $d(H_0)$ is the distance from any point on $L_2$ (e.g.,$(0,1,-1)$) to $H_0$: $d = \frac{|0 - (-1)|}{\sqrt{1^2 + (-1)^2}} = \frac{1}{\sqrt{2}}$. Thus,$P \rightarrow 5$.$(Q)$ Distance of $(0,1,2)$ from $x-z=0$ is $\frac{|0-2|}{\sqrt{1^2+(-1)^2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$. Thus,$Q \rightarrow 4$.$(R)$ Since $H_0$ is $x-z=0$,it passes through the origin $(0,0,0)$. Thus,the distance is $0$. $R \rightarrow 3$.$(S)$ Intersection of $y=z, x=1, x-z=0$: From $x=1$ and $x-z=0$,$z=1$. From $y=z$,$y=1$. Point is $(1,1,1)$. Distance from origin is $\sqrt{1^2+1^2+1^2} = \sqrt{3}$. Thus,$S \rightarrow 1$.Therefore,the correct matching is $(P) \rightarrow (5), (Q) \rightarrow (4), (R) \rightarrow (3), (S) \rightarrow (1)$.

List-$I$	List-$II$
$(P)$ The value of $d(H_0)$ is	$(1)$ $\sqrt{3}$
$(Q)$ The distance of the point $(0,1,2)$ from $H_0$ is	$(2)$ $\frac{1}{\sqrt{3}}$
$(R)$ The distance of origin from $H_0$ is	$(3)$ $0$
$(S)$ The distance of origin from the point of intersection of planes $y=z, x=1$ and $H_0$ is	$(4)$ $\sqrt{2}$
	$(5)$ $\frac{1}{\sqrt{2}}$

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