The equation of the plane which is equally inclined to the lines `(x-1)/2=y/(-2)=(z+2)/(-1)a n d=(x+3)/8=(y-4)/1=z/(-4)` and passing through the origin is/are a. `14 x-5y-7z=0` b. `2x+7y-z=0` c. `3x-4y-z=0` d. `x+2y-5z=0`

Options

A. `14x-5y-7z=0`
B. `2x+7y-z=0`
C. `3x-4y-z=0`
D. `x+2y-5z=0`

Explanation

To find the equation of the plane that is equally inclined to the given lines and passes through the origin, we can follow these steps:

### Step 1: Identify the direction ratios of the lines
The first line is given by:
\[ \frac{x-1}{2} = \frac{y}{-2} = \frac{z+2}{-1} \]
From this, we can extract the direction ratios as \( (2, -2, -1) \).

The second line is given by:
\[ \frac{x+3}{8} = \frac{y-4}{1} = \frac{z}{-4} \]
From this, we can extract the direction ratios as \( (8, 1, -4) \).

### Step 2: Convert direction ratios to vectors
Let \( \mathbf{a} = (2, -2, -1) \) and \( \mathbf{b} = (8, 1, -4) \).

### Step 3: Find the unit vectors of the direction ratios
To find the unit vectors, we first calculate the magnitudes of \( \mathbf{a} \) and \( \mathbf{b} \).

For \( \mathbf{a} \):
\[ |\mathbf{a}| = \sqrt{2^2 + (-2)^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \]
Thus, the unit vector \( \mathbf{u_a} \) is:
\[ \mathbf{u_a} = \left(\frac{2}{3}, \frac{-2}{3}, \frac{-1}{3}\right) \]

For \( \mathbf{b} \):
\[ |\mathbf{b}| = \sqrt{8^2 + 1^2 + (-4)^2} = \sqrt{64 + 1 + 16} = \sqrt{81} = 9 \]
Thus, the unit vector \( \mathbf{u_b} \) is:
\[ \mathbf{u_b} = \left(\frac{8}{9}, \frac{1}{9}, \frac{-4}{9}\right) \]

### Step 4: Find the equation of the plane
The equation of a plane that is equally inclined to the two lines can be represented using the sum of the unit vectors:
\[ \mathbf{n} = \mathbf{u_a} + \mathbf{u_b} \]
Calculating this gives:
\[ \mathbf{n} = \left(\frac{2}{3} + \frac{8}{9}, \frac{-2}{3} + \frac{1}{9}, \frac{-1}{3} + \frac{-4}{9}\right) \]

To add these fractions, we convert them to a common denominator:
- For \( \frac{2}{3} \) we convert to \( \frac{6}{9} \):
\[ \mathbf{n} = \left(\frac{6}{9} + \frac{8}{9}, \frac{-6}{9} + \frac{1}{9}, \frac{-3}{9} + \frac{-4}{9}\right) = \left(\frac{14}{9}, \frac{-5}{9}, \frac{-7}{9}\right) \]

### Step 5: Write the equation of the plane
The normal vector \( \mathbf{n} \) gives us the coefficients of the plane equation:
\[ 14x - 5y - 7z = 0 \]

### Step 6: Check for the second case (subtracting the unit vectors)
Now, we also consider the case where we subtract the unit vectors:
\[ \mathbf{n'} = \mathbf{u_a} - \mathbf{u_b} \]
Calculating this gives:
\[ \mathbf{n'} = \left(\frac{2}{3} - \frac{8}{9}, \frac{-2}{3} - \frac{1}{9}, \frac{-1}{3} - \frac{-4}{9}\right) \]
Converting to a common denominator:
\[ \mathbf{n'} = \left(\frac{6}{9} - \frac{8}{9}, \frac{-6}{9} - \frac{1}{9}, \frac{-3}{9} + \frac{4}{9}\right) = \left(\frac{-2}{9}, \frac{-7}{9}, \frac{1}{9}\right) \]

### Step 7: Write the equation of the second plane
The equation of the plane is:
\[ -2x - 7y + z = 0 \quad \text{or} \quad 2x + 7y - z = 0 \]

### Final Result
Thus, the equations of the planes that are equally inclined to the given lines and pass through the origin are:
1. \( 14x - 5y - 7z = 0 \) (Option a)
2. \( 2x + 7y - z = 0 \) (Option b)