Let $l_{1}$ and $l_{2}$ be the lines ${\vec{r}}_{1} = λ (\hat{i} + \hat{j} + \hat{k})$ and ${\vec{r}}_{2} = (\hat{j} - \hat{k}) + μ (\hat{i} + \hat{k}),$ respectively. Let $X$ be the set of all the planes $H$ that contain the line $l_{1} .$ For a plane $H,$ let $d (H)$ denote the smallest possible distance between the points of $l_{2}$ 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 $X .$

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distance of $(0, 1, 2)$ from $H_{0}$ is $\sqrt{2}$

$d (H_{0}) = \frac{1}{\sqrt{2}}$

distance of origin from $H_{0}$ is 0

distance of origin from the point of intersection of $y = z, x = 1$ and $H_{0}$ is $\sqrt{3}$

answer is A, B, C, D.

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Detailed Solution

$l_{1} \vec{:} r_{1} = λ (i + j + k) = \bar{a} + λ \bar{b} l_{2} : \vec{r_{2}} = j - k + μ (i + k) = \bar{c} + μ \bar{d} H e r e, [\begin{matrix} \bar{a} - \bar{c} & \bar{b} & \bar{d} \end{matrix}] = | \begin{matrix} 0 & 1 & - 1 \\ 1 & 1 & 1 \\ 1 & 0 & 1 \end{matrix} | = 1 \neq 0$

$\Rightarrow l_{1}$ and $l_{2}$ are skew – lines.
$∴ H_{0}$ is the plane containing $l_{1}$ and parallel to $l_{2}$
$\Rightarrow$ Equation of $H_{2}$ is $| \begin{matrix} x & y & z \\ 1 & 1 & 1 \\ 1 & 0 & 1 \end{matrix} | = 0 \Rightarrow x - z = 0$
P) $d (H_{0}) =$ Shortest distance between $d (H_{0}) =$ and $l_{2} = \frac{1}{\sqrt{2}}$
Q) $d = | \frac{0 - 2}{\sqrt{2}} | = \sqrt{2}$
R) $d = | \frac{0 - 0}{\sqrt{2}} | = 0$
$H_{0}$ passes through $(0, 0, 0)$
S) Point of intersection of given planes and $H_{0}$ is $(1, 1, 1)$
Distance from origin = $\sqrt{{(1 - 0)}^{2} + {(1 - 0)}^{2} + {(1 - 0)}^{2}} = \sqrt{3}$