The equation of the line passing through `(1, 1, 1)` and perpendicular to the line of intersection of the planes `x+2y-4z=0 and 2x-y+2z=0` is

Options

A. `(x-1)/(5)=(1-y)/(1)=(z-1)/(2)`
B. `(x-1)/(-5)=(1-y)/(1)=(z-1)/(2)`
C. `(x-1)/(0)=(1-y)/(-10)=(z-1)/(-5)`
D. `(x-1)/(-10)=(y+2)/(0)=(z-2)/(-5)`

Answer

To find the equation of the line passing through the point \( (1, 1, 1) \) and perpendicular to the line of intersection of the planes given by the equations \( x + 2y - 4z = 0 \) and \( 2x - y + 2z = 0 \), we can follow these steps:

### Step 1: Find the Normal Vectors of the Planes
The normal vector of a plane given by the equation \( ax + by + cz = d \) is \( (a, b, c) \).

For the first plane \( x + 2y - 4z = 0 \):
- The normal vector \( \mathbf{n_1} = (1, 2, -4) \).

For the second plane \( 2x - y + 2z = 0 \):
- The normal vector \( \mathbf{n_2} = (2, -1, 2) \).

### Step 2: Find the Direction Vector of the Line of Intersection
The direction vector \( \mathbf{d} \) of the line of intersection of the two planes can be found using the cross product of the normal vectors \( \mathbf{n_1} \) and \( \mathbf{n_2} \).

\[ \mathbf{d} = \mathbf{n_1} \times \mathbf{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -4 \\ 2 & -1 & 2 \end{vmatrix} \]

Calculating the determinant:

\[ \mathbf{d} = \mathbf{i} \left( 2 \cdot 2 - (-4) \cdot (-1) \right) - \mathbf{j} \left( 1 \cdot 2 - (-4) \cdot 2 \right) + \mathbf{k} \left( 1 \cdot (-1) - 2 \cdot 2 \right) \]

\[ = \mathbf{i} (4 - 4) - \mathbf{j} (2 + 8) + \mathbf{k} (-1 - 4) \]

\[ = \mathbf{i} (0) - \mathbf{j} (10) + \mathbf{k} (-5) \]

Thus, the direction vector \( \mathbf{d} = (0, -10, -5) \).

### Step 3: Find the Direction Ratios of the Perpendicular Line
The line we want to find is perpendicular to the line of intersection, which means its direction ratios \( (l, m, n) \) must satisfy the dot product condition with the direction vector \( (0, -10, -5) \):

\[ 0 \cdot l + (-10) \cdot m + (-5) \cdot n = 0 \]

This simplifies to:

\[ -10m - 5n = 0 \implies 2m + n = 0 \implies n = -2m \]

### Step 4: Choose a Value for \( m \)
Let’s choose \( m = 1 \). Then:

\[ n = -2(1) = -2 \]

Thus, the direction ratios of the required line can be taken as \( (1, 1, -2) \).

### Step 5: Write the Equation of the Line
The equation of a line in space can be expressed in symmetric form as:

\[ \frac{x - x_0}{l} = \frac{y - y_0}{m} = \frac{z - z_0}{n} \]

Substituting \( (x_0, y_0, z_0) = (1, 1, 1) \) and \( (l, m, n) = (1, 1, -2) \):

\[ \frac{x - 1}{1} = \frac{y - 1}{1} = \frac{z - 1}{-2} \]

### Final Answer
The equation of the line is:

\[ x - 1 = y - 1 = \frac{1 - z}{2} \]