cosider the lines,
`L1=(x+1)/3=(y+2)/1=(z+2)/2`
`L2=(x-2)/1=(y+2)/2=(z-3)/3`
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)`
B. `7/sqrt(75)`
C. `13/sqrt(75)`
D. `23/sqrt(75)`

Answer

To solve the problem, we need to find the distance of the point \( (1, 1, 1) \) from a plane that passes through the point \( (-1, -2, -1) \) and has a normal vector that is perpendicular to both lines \( L_1 \) and \( L_2 \).

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

The lines are given in symmetric form:

- For line \( L_1 \):
\[ \frac{x + 1}{3} = \frac{y + 2}{1} = \frac{z + 2}{2} \]
The direction vector \( \mathbf{d_1} \) of \( L_1 \) is \( (3, 1, 2) \).

- For line \( L_2 \):
\[ \frac{x - 2}{1} = \frac{y + 2}{2} = \frac{z - 3}{3} \]
The direction vector \( \mathbf{d_2} \) of \( L_2 \) is \( (1, 2, 3) \).

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

The normal vector \( \mathbf{n} \) to the plane can be found by taking the cross product of the direction vectors \( \mathbf{d_1} \) and \( \mathbf{d_2} \):
\[ \mathbf{d_1} = (3, 1, 2), \quad \mathbf{d_2} = (1, 2, 3) \]

Calculating the cross product:
\[ \mathbf{n} = \mathbf{d_1} \times \mathbf{d_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 1 & 2 \\ 1 & 2 & 3 \end{vmatrix} \]

Expanding the determinant:
\[ \mathbf{n} = \mathbf{i}(1 \cdot 3 - 2 \cdot 2) - \mathbf{j}(3 \cdot 3 - 2 \cdot 1) + \mathbf{k}(3 \cdot 2 - 1 \cdot 1) \]
\[ = \mathbf{i}(3 - 4) - \mathbf{j}(9 - 2) + \mathbf{k}(6 - 1) \]
\[ = -\mathbf{i} - 7\mathbf{j} + 5\mathbf{k} \]
Thus, the normal vector \( \mathbf{n} = (-1, -7, 5) \).

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

The equation of a plane can be expressed as:
\[ n_1(x - x_0) + n_2(y - y_0) + n_3(z - z_0) = 0 \]
where \( (x_0, y_0, z_0) \) is a point on the plane and \( (n_1, n_2, n_3) \) is the normal vector.

Using the point \( (-1, -2, -1) \) and the normal vector \( (-1, -7, 5) \):
\[ -1(x + 1) - 7(y + 2) + 5(z + 1) = 0 \]
Expanding this:
\[ -x - 1 - 7y - 14 + 5z + 5 = 0 \]
\[ -x - 7y + 5z - 10 = 0 \]
Rearranging gives:
\[ x + 7y - 5z + 10 = 0 \]

### Step 4: Calculate the distance from the point \( (1, 1, 1) \) to the plane

The distance \( D \) from a point \( (x_1, y_1, z_1) \) to the plane \( ax + by + cz + d = 0 \) is given by:
\[ D = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}} \]

Here, \( a = 1, b = 7, c = -5, d = 10 \) and the point is \( (1, 1, 1) \):
\[ D = \frac{|1 \cdot 1 + 7 \cdot 1 - 5 \cdot 1 + 10|}{\sqrt{1^2 + 7^2 + (-5)^2}} \]
Calculating the numerator:
\[ = |1 + 7 - 5 + 10| = |13| = 13 \]
Calculating the denominator:
\[ = \sqrt{1 + 49 + 25} = \sqrt{75} \]
Thus, the distance is:
\[ D = \frac{13}{\sqrt{75}} = \frac{13}{5\sqrt{3}} = \frac{13\sqrt{3}}{15} \]

### Final Answer
The distance of the point \( (1, 1, 1) \) from the plane is \( \frac{13}{\sqrt{75}} \).