Let ℓ₁ and ℓ₂ be the lines

$r_1=\lambda (i+j+k)$ and $r_2=(j-k)+\mu (i+k)$,

respectively. Let X be the set of all the planes H that contain the line ℓ₁. For a plane H, let d(H) denote the smallest possible distance between the points of ℓ₂ and H. Let H₀ be a plane in X for which d(H₀) 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) $\text{The value of}d\left(H_0\right)\text{is}$	(1) $\sqrt{3}$
(Q) $\text{The distance of the point}(0,1,2)\text{from}{H}_0\text{is}$	(2) $\frac{1}{\sqrt{3}}$
(R) $\text{The distance of origin from}{H}_0\text{is}$	(3) $0$
(S) $\text{The distance of origin from point of intersection of planes}y=z,x=1\text{and}{H}_0\text{is}$	(4) $\sqrt{2}$
	(5) $\frac{1}{\sqrt{2}}$

The correct option is (P) → (5), (Q) → (4), (R) → (3), (S) → (1).

Step 1: Understanding the geometry of the lines and planes

Line ℓ₁ passes through the origin (0, 0, 0) and has direction vector $b_1=\hat{i}+\hat{j}+\hat{k}$.

Line ℓ₂ passes through point A(0, 1, -1) and has direction vector $b_2=\hat{i}+0\hat{j}+\hat{k}$.

Let H be any plane containing the line ℓ₁. Since ℓ₁ contains the origin (0, 0, 0), the equation of plane H must be of the form:

$ax+by+cz=0$

where the normal vector $\hat{n}=(a,b,c)$ is perpendicular to line ℓ₁, giving:

$a+b+c=0\Rightarrow c=-(a+b)$

Step 2: Finding d(H)

The distance d(H) between line ℓ₂ and plane H is non-zero only when ℓ₂ is parallel to H. If ℓ₂ is not parallel to H, ℓ₂ intersects H, making the distance d(H) = 0.

To maximize d(H), plane H must be parallel to line ℓ₂. Thus, the normal vector $\hat{n}$ must be perpendicular to $b_2$:

$\hat{n}·b_2=0\Rightarrow a\left(1\right)+b\left(0\right)+c\left(1\right)=0\Rightarrow a+c=0$

Since $c=-(a+b)$, substituting this gives:

$a-a-b=0\Rightarrow b=0$

This implies $c=-a$. Therefore, the normal vector to H₀ is proportional to $(1,0,-1)$.

Thus, the equation of plane H₀ is:

$x-z=0$

Step 3: Matching List-I to List-II

(P) Value of d(H₀):

The distance d(H₀) is the perpendicular distance from point A(0, 1, -1) on ℓ₂ to the plane H₀ ($x-z=0$):

$d\left(H_0\right)=\frac{|0-(-1\left)\right|}{\sqrt{1^2+0^2+{(-1)}^2}}=\frac{1}{\sqrt{2}}$

Hence, (P) → (5).

(Q) Distance of point (0, 1, 2) from H₀:

$\text{Distance}=\frac{|0-2|}{\sqrt{1^2+0^2+{(-1)}^2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$

Hence, (Q) → (4).

(R) Distance of origin from H₀:

Since H₀ is $x-z=0$, it passes directly through the origin (0, 0, 0). Thus, the distance is 0.

Hence, (R) → (3).

(S) Distance of origin from point of intersection of planes y = z, x = 1, and H₀:

Solving the system of equations:

1) $x=1$
2) $x-z=0\Rightarrow z=1$
3) $y=z\Rightarrow y=1$

The point of intersection is (1, 1, 1).

The distance of (1, 1, 1) from origin (0, 0, 0) is:

$\sqrt{1^2+1^2+1^2}=\sqrt{3}$

Hence, (S) → (1).