Given lines

$\ell_1: \vec r=\lambda(\hat i+\hat j+\hat k)$ so $\ell_1$ passes through origin with direction vector $\vec a=(1,1,1).$

$\ell_2:(x,y,z)=(0,1,-1)+\mu(1,0,1)$ since $\hat j-\hat k=(0,1,-1)$ and direction vector is $\vec b=(1,0,1).$

General plane containing $\ell_1$

Any plane $H\in X$ containing $\ell_1$ must pass through the origin and contain direction $(1,1,1)$. Hence its normal vector $\vec n$ satisfies $\vec n\cdot (1,1,1)=0.$ Therefore plane equation is $\vec n\cdot \vec r=0,$ where $\vec n\perp (1,1,1).$$

Distance from line $\ell_2$ to plane $H$

A point on $\ell_2$ is $\vec r=(0,1,-1)+\mu(1,0,1).$ Its signed value in the plane equation is $\vec n\cdot \vec r=\vec n\cdot(0,1,-1)+\mu\,\vec n\cdot(1,0,1).$

The minimum distance from points of $\ell_2$ to plane $H$ is zero if the line intersects the plane, i.e. if $\vec n\cdot(1,0,1)\neq 0$.

For a positive distance, $\ell_2$ must be parallel to the plane, so we need $\vec n\cdot(1,0,1)=0.$

Thus $\vec n$ must be perpendicular to both $(1,1,1)$ and $(1,0,1)$. Hence $\vec n \parallel (1,1,1)\times(1,0,1).$

Compute:

So $H_0$ has normal vector $(1,0,-1)$, hence equation $x-z=0.$

Compute $d(H_0)$

Since $\ell_2$ is parallel to $H_0$, distance from the line to plane equals distance of any point on $\ell_2$ from plane. Take point $(0,1,-1)$.

Distance from $(x_0,y_0,z_0)$ to plane $x-z=0$ is

Therefore $d(H_0)=\frac{|0-(-1)|}{\sqrt2}=\frac{1}{\sqrt2}.$

So $(P)\to(5).$

Distance of point $(0,1,2)$ from $H_0$

$\text{dist}((0,1,2),H_0)=\frac{|0-2|}{\sqrt2}=\frac{2}{\sqrt2}=\sqrt2.$

So $(Q)\to(4).$

Distance of origin from $H_0$

Since plane $H_0: x-z=0$ passes through origin, $\text{dist}((0,0,0),H_0)=0.$

So $(R)\to(3).$

Intersection of planes $y=z$, $x=1$, and $H_0$

From $H_0$, we have $x=z.$ Also given $x=1,\qquad y=z.$ Hence $x=1,\ z=1,\ y=1.$ So intersection point is $(1,1,1).$

Distance of origin from this point is $\sqrt{1^2+1^2+1^2}=\sqrt3.$

Thus $(S)\to(1).$

Final matching

This corresponds to Option B.