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 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 a 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**
(P) The value of $d(H_0)$ is
(Q) The distance of the point $(0,1,2)$ from $H_0$ is
(R) The distance of the origin from $H_0$ is
(S) The distance of the origin from the point of intersection of the planes $y=z$, $x=1$ and $H_0$ is
**List-II**
(1) $\sqrt{3}$
(2) $\dfrac{1}{\sqrt{3}}$
(3) 0
(4) $\sqrt{2}$
(5) $\dfrac{1}{\sqrt{2}}$
(P)→(2) (Q)→(4) (R)→(5) (S)→(1)
(P)→(5) (Q)→(4) (R)→(3) (S)→(1)
(P)→(2) (Q)→(1) (R)→(3) (S)→(2)
(P)→(5) (Q)→(1) (R)→(4) (S)→(2)
Step-by-Step Solution
Key Concept: d(H) is nonzero only when ℓ₂ is parallel to H; two perpendicularity conditions pin down n uniquely
$\ell_1$: through origin, direction $\vec{d_1}=(1,1,1)$. Plane $H\in X$ has normal $\vec{n}\perp(1,1,1)$ and passes through origin.
$d(H)=0$ if $\ell_2$ meets $H$ (i.e., $(\hat{i}+\hat{k})\cdot\vec{n}\neq0$). To maximise $d(H)$, need $\ell_2\parallel H$: $(1,0,1)\cdot\vec{n}=0$.
Combine with $n_1+n_2+n_3=0$ and $n_1+n_3=0\Rightarrow n_2=0$. So $\vec{n}=(1,0,-1)/\sqrt{2}$.
$H_0$: plane $x-z=0$ through origin.
(P) $d(H_0)=|(0,1,-1)\cdot(1,0,-1)/\sqrt{2}|=|0+0+1|/\sqrt{2}=1/\sqrt{2}$→(5).
(Q) Distance of $(0,1,2)$ from $H_0$ ($x-z=0$): $|0-2|/\sqrt{2}=2/\sqrt{2}=\sqrt{2}$→(4).
(R) Distance of origin from $H_0$: $H_0$ passes through origin → 0→(3).
(S) Intersection of $y=z$, $x=1$, $H_0$ ($x=z$): from $x=1$ and $x=z$: $z=1$, from $y=z$: $y=1$. Point $(1,1,1)$. Distance from origin $=\sqrt{3}$→(1).
Answer: (P)→(5),(Q)→(4),(R)→(3),(S)→(1) → B.
Correct Answer: B