Let $$\ell_1$$ and $$\ell_2$$ be the lines $$\vec{r}_1 = \lambda(\hat{i} + \hat{j} + \hat{k})$$ and $$\vec{r}_2 = (\hat{j} - \hat{k}) + \mu(\hat{i} + \hat{k})$$, respectively. Let X be the set of all the planes H that contain the line $$\ell_1$$. For a plane H, let d(H) denote the smallest possible distance between the points of $$\ell_2$$ and H. Let $$H_0$$ be the plane in X for which d($$H_0$$) is the maximum value of d(H) as H varies over all planes in X.

Match each entry in List-I to the correct entries in List-II.

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)	$$\frac{1}{\sqrt{2}}$$

The correct option is:

The two lines are
$$\ell_1: \;\vec r = \lambda(\hat i+\hat j+\hat k),\;\lambda\in\mathbb R$$ and
$$\ell_2: \;\vec r = (\hat j-\hat k)+\mu(\hat i+\hat k),\;\mu\in\mathbb R.$$

1. General form of a plane through $$\ell_1$$
Any plane $$H$$ containing $$\ell_1$$ must pass through the origin (because $$\ell_1$$ does) and must contain the direction vector $$\langle1,1,1\rangle$$.
If $$\mathbf n=\langle a,b,c\rangle$$ is the normal to $$H$$, then $$\mathbf n\cdot\langle1,1,1\rangle=0\;$$ i.e.
$$a+b+c=0\;-(1)$$
Hence every such plane has equation
$$ax+by+cz=0,\qquad a+b+c=0.$$\

2. Distance of a point on $$\ell_2$$ from $$H$$
A general point on $$\ell_2$$ is
$$P(\mu,\,1,\,-1+\mu).$$
Its perpendicular distance from $$H$$ is
$$d(\mu)=\frac{\lvert a\mu+b+c(-1+\mu)\rvert}{\sqrt{a^{2}+b^{2}+c^{2}}}$$ $$=\frac{\lvert\mu(a+c)+b-c\rvert}{\sqrt{a^{2}+b^{2}+c^{2}}}\;-(2).$$

3. Least distance from $$\ell_2$$ to $$H$$
For a fixed plane $$H$$ (fixed $$a,b,c$$) we can vary $$\mu$$ to minimise the numerator in (2).
• If $$a+c\neq0$$ we choose $$\mu=\dfrac{a+2c}{a+c}$$ to make the numerator zero, giving the minimum distance 0.
• If $$a+c=0$$, then from (1) $$b=-(a+c)=0$$ and $$c=-a$$.   The numerator in (2) becomes $$\lvert-a-2c\rvert=\lvert a\rvert$$, independent of $$\mu$$, so the least distance is
$$d(H)=\frac{\lvert a\rvert}{\sqrt{a^{2}+0+(-a)^{2}}}=\frac{\lvert a\rvert}{\lvert a\rvert\sqrt2}=\frac1{\sqrt2}\;-(3).$$

4. Plane $$H_0$$ giving the maximum of the least distances
For most planes the least distance is 0; the largest possible least distance (from (3)) is $$\dfrac1{\sqrt2}$$, attained when $$a+c=0$$.
Taking the simplest normal $$\mathbf n=\langle1,0,-1\rangle$$ (i.e. $$a=1,\,b=0,\,c=-1$$) we obtain
$$H_0:\;x-z=0\;(x=z).$$

Case P: $$d(H_0)=\dfrac1{\sqrt2}\Rightarrow(P)\to(5).$

5. Distance of (0,1,2) from $$H_0$$
$$d=$$\frac{\lvert0-2\rvert}{\sqrt{1^{2}$$+0^{2}+(-1)^{2}}}=$$\frac{2}{\sqrt2}=\sqrt$$2$$\Rightarrow$$(Q)$$\to$$(4).$$

6. Distance of origin from $$H_0$$
The origin lies on $$H_0$$, so the distance is 0 → $$(R)$$\to$$(3).$$

7. Point common to $$y=z,\;x=1,$$ and $$H_0$$
From $$x=1,\;x=z$$ we get $$z=1$$; from $$y=z$$, $$y=1$$. Thus the intersection point is $$A(1,1,1)$$.
Distance OA $$=$$\sqrt{1^{2}$$+1^{2}+1^{2}}=$$\sqrt$$3$$\Rightarrow$$(S)$$\to$$(1).$$

8. Final matching
$$$$\begin{aligned}$$ (P)&$$\to$$(5),\\ (Q)&$$\to$$(4),\\ (R)&$$\to$$(3),\\ (S)&$$\to$$(1). \end{aligned}$$
Hence the correct option is Option B.