Let `L_(1)` be the line `vecr_(1)=2hati+hatj-hatk+lamda(hati+2hatk)` and let `L_(2)` be the line `vecr_(2)=3hati+hatj-hatk+mu(hati+hatj+hatk)`. Let `pi` be the plane which contains the line `L_(1)` and is parallel to `L_(2)`. The distance of the plane `pi` from the origin is

Options

A. `sqrt(2//7)`
B. `1//7`
C. `sqrt6`
D. none

Explanation

To find the distance of the plane \( \pi \) from the origin, we will follow these steps:

### Step 1: Identify the lines and their direction ratios
The lines are given as:
- \( L_1: \vec{r}_1 = 2\hat{i} + \hat{j} - \hat{k} + \lambda(\hat{i} + 2\hat{k}) \)
- \( L_2: \vec{r}_2 = 3\hat{i} + \hat{j} - \hat{k} + \mu(\hat{i} + \hat{j} + \hat{k}) \)

From \( L_1 \), the direction ratios are \( (1, 0, 2) \) (from \( \hat{i} + 2\hat{k} \)).
From \( L_2 \), the direction ratios are \( (1, 1, 1) \) (from \( \hat{i} + \hat{j} + \hat{k} \)).

### Step 2: Determine the normal vector of the plane
Since the plane \( \pi \) contains the line \( L_1 \) and is parallel to \( L_2 \), the normal vector \( \vec{n} \) of the plane can be found using the cross product of the direction ratios of \( L_1 \) and \( L_2 \).

Let:
- \( \vec{d_1} = (1, 0, 2) \)
- \( \vec{d_2} = (1, 1, 1) \)

The normal vector \( \vec{n} = \vec{d_1} \times \vec{d_2} \).

Calculating the cross product:
\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 2 \\ 1 & 1 & 1 \end{vmatrix} = \hat{i}(0 \cdot 1 - 2 \cdot 1) - \hat{j}(1 \cdot 1 - 2 \cdot 1) + \hat{k}(1 \cdot 1 - 0 \cdot 1) \]
\[ = -2\hat{i} + \hat{j} + \hat{k} \]

### Step 3: Use a point on line \( L_1 \) to find the equation of the plane
A point on line \( L_1 \) can be taken as \( (2, 1, -1) \).

The equation of the plane can be expressed as:
\[ -2(x - 2) + 1(y - 1) + 1(z + 1) = 0 \]
Expanding this:
\[ -2x + 4 + y - 1 + z + 1 = 0 \]
\[ -2x + y + z + 4 = 0 \]
or
\[ 2x - y - z = 4 \]

### Step 4: Find the distance from the origin to the plane
The distance \( d \) from a point \( (x_0, y_0, z_0) \) to the plane \( Ax + By + Cz + D = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
For the plane \( 2x - y - z - 4 = 0 \) (where \( A = 2, B = -1, C = -1, D = -4 \)), and the point \( (0, 0, 0) \):
\[ d = \frac{|2(0) - 1(0) - 1(0) - 4|}{\sqrt{2^2 + (-1)^2 + (-1)^2}} = \frac{|-4|}{\sqrt{4 + 1 + 1}} = \frac{4}{\sqrt{6}} = \frac{4\sqrt{6}}{6} = \frac{2\sqrt{6}}{3} \]

### Final Answer
The distance of the plane \( \pi \) from the origin is \( \frac{2\sqrt{6}}{3} \).

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