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

Options

A. `x-3y-2z=-2`
B. `2x -z=2`
C. `x-y-z=0`
D. `x+3y+ z=`

Answer

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

### Step 1: Write the equations of the given planes
The equations of the two planes are:
1. \(P_1: 2x - y - 4 = 0\)
2. \(P_2: y + 2z - 4 = 0\)

### Step 2: Use the formula for the equation of a plane containing the line of intersection
The equation of a plane that contains the line of intersection of the two planes can be expressed as:
\[ P_1 + \lambda P_2 = 0 \]
Substituting the equations of the planes, we have:
\[ (2x - y - 4) + \lambda (y + 2z - 4) = 0 \]

### Step 3: Expand the equation
Expanding this, we get:
\[ 2x - y - 4 + \lambda y + 2\lambda z - 4\lambda = 0 \]
Rearranging gives:
\[ 2x + (-1 + \lambda)y + 2\lambda z - (4 + 4\lambda) = 0 \]

### Step 4: Substitute the point (1, 1, 0) into the equation
Since the plane must pass through the point \((1, 1, 0)\), we substitute \(x = 1\), \(y = 1\), and \(z = 0\) into the equation:
\[ 2(1) + (-1 + \lambda)(1) + 2\lambda(0) - (4 + 4\lambda) = 0 \]
This simplifies to:
\[ 2 - 1 + \lambda - 4 - 4\lambda = 0 \]
Combining like terms results in:
\[ -3 - 3\lambda = 0 \]

### Step 5: Solve for \(\lambda\)
Solving for \(\lambda\):
\[ -3\lambda = 3 \implies \lambda = -1 \]

### Step 6: Substitute \(\lambda\) back into the plane equation
Now substitute \(\lambda = -1\) back into the plane equation:
\[ 2x + (-1 - 1)y + 2(-1)z - (4 - 4) = 0 \]
This simplifies to:
\[ 2x - 2y - 2z = 0 \]

### Step 7: Simplify the equation
Dividing the entire equation by 2 gives:
\[ x - y - z = 0 \]

### Final Answer
Thus, the equation of the required plane is:
\[ \boxed{x - y - z = 0} \]