The distance of the point `(1, 3, -7)` from the plane passing through the point `(1, -1, -1)` having normal perpendicular to both the lines `(x-1)/(1)=(y+2)/(-2)=(z-4)/(3) and (x-2)/(2)=(y+1)/(-1)=(z+7)/(-1)` is

Options

A. `(20)/(sqrt(74))` units
B. `(10)/(sqrt(83))` units
C. `(5)/(sqrt(83))` units
D. `(10)/(sqrt(74))` units

Answer

To find the distance of the point \( P(1, 3, -7) \) from the plane passing through the point \( A(1, -1, -1) \) with a normal vector perpendicular to the two given lines, we will follow these steps:

### Step 1: Identify the direction vectors of the lines
The equations of the two lines are given in symmetric form:
1. \( \frac{x-1}{1} = \frac{y+2}{-2} = \frac{z-4}{3} \)
2. \( \frac{x-2}{2} = \frac{y+1}{-1} = \frac{z+7}{-1} \)

From these equations, we can extract the direction vectors:
- For the first line, the direction vector \( \mathbf{n_1} = \langle 1, -2, 3 \rangle \)
- For the second line, the direction vector \( \mathbf{n_2} = \langle 2, -1, -1 \rangle \)

### Step 2: Find the normal vector to the plane
The normal vector \( \mathbf{N} \) to the plane can be found using the cross product of the two direction vectors \( \mathbf{n_1} \) and \( \mathbf{n_2} \).

\[ \mathbf{N} = \mathbf{n_1} \times \mathbf{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -2 & 3 \\ 2 & -1 & -1 \end{vmatrix} \]

Calculating the determinant:

\[ \mathbf{N} = \mathbf{i}((-2)(-1) - (3)(-1)) - \mathbf{j}((1)(-1) - (3)(2)) + \mathbf{k}((1)(-1) - (-2)(2)) \]

\[ = \mathbf{i}(2 + 3) - \mathbf{j}(-1 - 6) + \mathbf{k}(-1 + 4) \]

\[ = 5\mathbf{i} + 7\mathbf{j} + 3\mathbf{k} \]

Thus, the normal vector is \( \mathbf{N} = \langle 5, 7, 3 \rangle \).

### Step 3: Write the equation of the plane
The equation of the plane can be expressed using the point-normal form. The plane passes through point \( A(1, -1, -1) \) and has the normal vector \( \langle 5, 7, 3 \rangle \).

The equation of the plane is given by:

\[ 5(x - 1) + 7(y + 1) + 3(z + 1) = 0 \]

Expanding this:

\[ 5x - 5 + 7y + 7 + 3z + 3 = 0 \]

\[ 5x + 7y + 3z + 5 = 0 \]

### Step 4: Calculate the distance from the point to the plane
The distance \( d \) from a point \( P(x_0, y_0, z_0) \) to the plane \( Ax + By + Cz + D = 0 \) is given by the formula:

\[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]

Here, \( A = 5 \), \( B = 7 \), \( C = 3 \), \( D = 5 \), and the point \( P(1, 3, -7) \):

Calculating:

\[ d = \frac{|5(1) + 7(3) + 3(-7) + 5|}{\sqrt{5^2 + 7^2 + 3^2}} \]

Calculating the numerator:

\[ = |5 + 21 - 21 + 5| = |10| = 10 \]

Calculating the denominator:

\[ = \sqrt{25 + 49 + 9} = \sqrt{83} \]

Thus, the distance is:

\[ d = \frac{10}{\sqrt{83}} \]

### Final Answer
The distance of the point \( (1, 3, -7) \) from the plane is \( \frac{10}{\sqrt{83}} \).

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