JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Three Dimensional Geometry

Let ${\ell }_1$ and ${\ell }_2$ be the lines ${\overrightarrow{\mathrm{r}}}_1=\lambda (\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})$ and ${\overrightarrow{\mathrm{r}}}_2=(\hat{\mathrm{j}}-\hat{\mathrm{k}})+\mu (\hat{\mathrm{i}}+\hat{\mathrm{k}})$, respectively. Let X be the set of all the planes H that contain the line ${\ell }_1$. For a plane H , let $\mathrm{d}(\mathrm{H})$ denote the smallest possible distance between the points of ${\ell }_2$ and H. Let ${\mathrm{H}}_0$ be plane in X for which $\mathrm{d}\left({\mathrm{H}}_0\right)$ is the maximum value of $\mathrm{d}(\mathrm{H})$ 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 $\mathrm{d}\left({\mathrm{H}}_0\right)$ is
(Q) The distance of the point (0,1,2) from ${\mathrm{H}}_0$ is
(R) The distance of origin from ${\mathrm{H}}_0$ is
(S) The distance of origin from the point of intersection of planes $\mathrm{y}=\mathrm{z},\mathrm{x}=1$ and ${\mathrm{H}}_0$ is

List-II
(1) $\sqrt{3}$
(2) $\frac{1}{\sqrt{3}}$
(3) 0
(4) $\sqrt{2}$
(5) $\frac{1}{\sqrt{2}}$

The correct option is :