Distance Properties of Planes Containing Line in Three Dimensional Geometry

To determine the correct matching between List-I and List-II, let us analyze the given lines and planes step-by-step.

1. Finding the Plane $H_0$

The given lines are:

• $\ell_1: \vec{r}_1 = \lambda(\hat{i} + \hat{j} + \hat{k})$, which passes through the origin $O(0,0,0)$ and has direction vector $\vec{b}_1 = \hat{i} + \hat{j} + \hat{k} = (1, 1, 1)$.
• $\ell_2: \vec{r}_2 = (\hat{j} - \hat{k}) + \mu(\hat{i} + \hat{k})$, which passes through $A(0, 1, -1)$ and has direction vector $\vec{b}_2 = \hat{i} + \hat{k} = (1, 0, 1)$.

Let $H$ be a plane containing $\ell_1$. Since $H$ contains $\ell_1$, it passes through the origin $O(0,0,0)$ and its normal vector $\vec{n} = (a, b, c)$ is perpendicular to $\vec{b}_1$: $\vec{n} \cdot \vec{b}_1 = a + b + c = 0 \implies c = -a - b$

The equation of any plane $H \in X$ is: $ax + by - (a + b)z = 0$

For any plane $H$, $d(H)$ represents the minimum distance between points of $\ell_2$ and $H$:

• If $\ell_2$ is not parallel to $H$, then $\ell_2$ intersects $H$, so $d(H) = 0$.
• If $\ell_2$ is parallel to $H$, then $d(H) > 0$, which is the distance from any point on $\ell_2$ to $H$.

To maximize $d(H)$, $H_0$ must be parallel to $\ell_2$. Therefore, the normal vector $\vec{n}$ must also be perpendicular to $\vec{b}_2$: $\vec{n} \cdot \vec{b}_2 = a(1) + b(0) + c(1) = a + c = 0$

Substituting $c = -a - b$: $a + (-a - b) = 0 \implies b = 0$

Thus, the normal vector to $H_0$ is proportional to: $\vec{n}_0 = (1, 0, -1)$

Since $H_0$ passes through the origin, its equation is: $1(x - 0) + 0(y - 0) - 1(z - 0) = 0 \implies x - z = 0$

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2. Evaluating the Entries in List-I

• Entry (P): The value of $d(H_0)$ Since $\ell_2$ is parallel to $H_0$, $d(H_0)$ is the perpendicular distance from the point $A(0,1,-1)$ on $\ell_2$ to the plane $H_0: x - z = 0$: $d(H_0) = \frac{|0 - (-1)|}{\sqrt{1^2 + 0^2 + (-1)^2}} = \frac{1}{\sqrt{2}}$ Hence, $(P) \rightarrow (5)$.

• Entry (Q): The distance of the point $(0,1,2)$ from $H_0$ Using the distance formula from a point $(x_1, y_1, z_1)$ to $x - z = 0$: $\text{Distance} = \frac{|0 - 2|}{\sqrt{1^2 + 0^2 + (-1)^2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$ Hence, $(Q) \rightarrow (4)$.

• Entry (R): The distance of the origin from $H_0$ The origin $(0,0,0)$ satisfies $x - z = 0 - 0 = 0$, so it lies on $H_0$. Thus, the distance is: $\text{Distance} = 0$ Hence, $(R) \rightarrow (3)$.

• Entry (S): The distance of origin from the point of intersection of planes $y = z$, $x = 1$, and $H_0$ To find the point of intersection, solve the system of equations:

  1. $x = 1$
  2. $x - z = 0 \implies z = x = 1$
  3. $y = z \implies y = 1$

  The point of intersection is $P(1, 1, 1)$. The distance of $P$ from the origin $(0,0,0)$ is: $\text{Distance} = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$ Hence, $(S) \rightarrow (1)$.

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Conclusion

Matching the results: $(P) \rightarrow (5), \quad (Q) \rightarrow (4), \quad (R) \rightarrow (3), \quad (S) \rightarrow (1)$

This corresponds to Option B.