Read the following passage and answer the questions. Consider the lines
`L_(1) : (x+1)/(3)=(y+2)/(1)=(z+1)/(2)`
`L_(2) : (x-2)/(1)=(y+2)/(2)=(z-3)/(3)`
Q. The distance of the point `(1, 1, 1)` from the plane passing through the point `(-1, -2, -1)` and whose normal is perpendicular to both the lines `L_(1) and L_(2)`, is

Options

A. `(2)/(sqrt(75))` unit
B. `(7)/(sqrt(75))` units
C. `(13)/(sqrt(75))` unit
D. `(23)/(sqrt(75))` units

Answer

To find the distance of the point \( (1, 1, 1) \) from the plane passing through the point \( (-1, -2, -1) \) and whose normal is perpendicular to both the lines \( L_1 \) and \( L_2 \), we will follow these steps:

### Step 1: Determine the direction ratios of the lines \( L_1 \) and \( L_2 \)

The equations of the lines are given as:
- \( L_1 : \frac{x+1}{3} = \frac{y+2}{1} = \frac{z+1}{2} \)
- \( L_2 : \frac{x-2}{1} = \frac{y+2}{2} = \frac{z-3}{3} \)

From these equations, we can extract the direction ratios:
- For \( L_1 \), the direction ratios are \( (3, 1, 2) \).
- For \( L_2 \), the direction ratios are \( (1, 2, 3) \).

### Step 2: Find the normal vector to the plane

The normal vector to the plane can be found by taking the cross product of the direction ratios of the two lines.

Let:
\[ \vec{d_1} = (3, 1, 2), \quad \vec{d_2} = (1, 2, 3) \]

The cross product \( \vec{n} = \vec{d_1} \times \vec{d_2} \) is calculated as follows:

\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ 1 & 2 & 3 \end{vmatrix} \]

Calculating the determinant:
\[ \vec{n} = \hat{i}(1 \cdot 3 - 2 \cdot 2) - \hat{j}(3 \cdot 3 - 2 \cdot 1) + \hat{k}(3 \cdot 2 - 1 \cdot 1) \]
\[ = \hat{i}(3 - 4) - \hat{j}(9 - 2) + \hat{k}(6 - 1) \]
\[ = \hat{i}(-1) - \hat{j}(7) + \hat{k}(5) \]
\[ = (-1, -7, 5) \]

### Step 3: Write the equation of the plane

The equation of a plane passing through a point \( (x_1, y_1, z_1) \) with a normal vector \( (a, b, c) \) is given by:
\[ a(x - x_1) + b(y - y_1) + c(z - z_1) = 0 \]

Substituting \( (x_1, y_1, z_1) = (-1, -2, -1) \) and \( (a, b, c) = (-1, -7, 5) \):
\[ -1(x + 1) - 7(y + 2) + 5(z + 1) = 0 \]
Expanding this:
\[ -x - 1 - 7y - 14 + 5z + 5 = 0 \]
\[ -x - 7y + 5z - 10 = 0 \]
Rearranging gives us:
\[ x + 7y - 5z = 10 \]

### Step 4: Calculate the distance from the point to the plane

The distance \( d \) from a point \( (x_0, y_0, z_0) \) to the plane \( ax + by + cz = d \) is given by:
\[ d = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}} \]

Substituting \( (x_0, y_0, z_0) = (1, 1, 1) \) and the coefficients from the plane equation \( (1, 7, -5) \) and \( d = 10 \):
\[ d = \frac{|1 \cdot 1 + 7 \cdot 1 - 5 \cdot 1 - 10|}{\sqrt{1^2 + 7^2 + (-5)^2}} \]
Calculating the numerator:
\[ = |1 + 7 - 5 - 10| = |-7| = 7 \]
Calculating the denominator:
\[ = \sqrt{1 + 49 + 25} = \sqrt{75} = 5\sqrt{3} \]
Thus, the distance is:
\[ d = \frac{7}{5\sqrt{3}} = \frac{7\sqrt{3}}{15} \]

### Final Answer
The distance of the point \( (1, 1, 1) \) from the plane is \( \frac{7\sqrt{3}}{15} \).

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