Equation of the line through the point `(1, 1, 1)` and intersecting the lines `2x-y-z-2=0=x+y+z-1 and x-y-z-3=0=2x+4y-z-4`.

Options

A. `x-1=0, 7x+17y-3z-134=0`
B. `x-1=0, 9x+15y-5z-19=0`
C. `x-1=0, (y-1)/(1)=(z-1)/(3)`
D. `x-2y+2z-1=0, 9x+15y-5z-19=0`

Answer

To find the equation of the line through the point \( (1, 1, 1) \) that intersects the given lines, we will follow these steps:

### Step 1: Identify the equations of the lines
We have two pairs of lines defined by the equations:
1. Line \( l_1: 2x - y - z - 2 = 0 \)
2. Line \( l_2: x + y + z - 1 = 0 \)

And,
3. Line \( l_3: x - y - z - 3 = 0 \)
4. Line \( l_4: 2x + 4y - z - 4 = 0 \)

### Step 2: Find the equation of the line intersecting \( l_1 \) and \( l_2 \)
To find a line that intersects both \( l_1 \) and \( l_2 \), we can express it as:
\[ l_1 + \lambda l_2 = 0 \]
Substituting the equations:
\[ (2x - y - z - 2) + \lambda (x + y + z - 1) = 0 \]
Expanding this gives:
\[ 2x - y - z - 2 + \lambda x + \lambda y + \lambda z - \lambda = 0 \]
Rearranging terms:
\[ (2 + \lambda)x + (-1 + \lambda)y + (-1 + \lambda)z - (2 + \lambda) = 0 \]

### Step 3: Substitute the point \( (1, 1, 1) \)
Since the line passes through the point \( (1, 1, 1) \), we substitute \( x = 1, y = 1, z = 1 \):
\[ (2 + \lambda)(1) + (-1 + \lambda)(1) + (-1 + \lambda)(1) - (2 + \lambda) = 0 \]
This simplifies to:
\[ (2 + \lambda) - (1 - \lambda) - (1 - \lambda) - (2 + \lambda) = 0 \]
Combining like terms:
\[ -2 + 2\lambda = 0 \]
Solving for \( \lambda \):
\[ \lambda = 1 \]

### Step 4: Substitute \( \lambda \) back into the equation
Substituting \( \lambda = 1 \) back into the equation:
\[ (2 + 1)x + (-1 + 1)y + (-1 + 1)z - (2 + 1) = 0 \]
This simplifies to:
\[ 3x - 3 = 0 \implies x - 1 = 0 \]

### Step 5: Find the equation of the line intersecting \( l_3 \) and \( l_4 \)
Now, we find the line that intersects \( l_3 \) and \( l_4 \):
\[ l_3 + \mu l_4 = 0 \]
Substituting the equations:
\[ (x - y - z - 3) + \mu (2x + 4y - z - 4) = 0 \]
Expanding gives:
\[ x - y - z - 3 + \mu(2x + 4y - z - 4) = 0 \]
Rearranging:
\[ (1 + 2\mu)x + (-1 + 4\mu)y + (-1 + \mu)z - (3 + 4\mu) = 0 \]

### Step 6: Substitute the point \( (1, 1, 1) \)
Substituting \( x = 1, y = 1, z = 1 \):
\[ (1 + 2\mu)(1) + (-1 + 4\mu)(1) + (-1 + \mu)(1) - (3 + 4\mu) = 0 \]
This simplifies to:
\[ (1 + 2\mu) - (1 - 4\mu) - (1 - \mu) - (3 + 4\mu) = 0 \]
Combining like terms:
\[ -4 + 7\mu = 0 \]
Solving for \( \mu \):
\[ \mu = \frac{4}{7} \]

### Step 7: Substitute \( \mu \) back into the equation
Substituting \( \mu = \frac{4}{7} \):
\[ (1 + 2 \cdot \frac{4}{7})x + (-1 + 4 \cdot \frac{4}{7})y + (-1 + \frac{4}{7})z - (3 + 4 \cdot \frac{4}{7}) = 0 \]
This simplifies to:
\[ \frac{15}{7}x + \frac{9}{7}y - \frac{3}{7}z - \frac{19}{7} = 0 \]
Multiplying through by 7 gives:
\[ 15x + 9y - 3z - 19 = 0 \]

### Final Result
The equations of the lines are:
1. \( x - 1 = 0 \)
2. \( 15x + 9y - 3z - 19 = 0 \)