Step 1: Understanding the Concept:
The equation of any plane passing through the line of intersection of two given planes \(P_1 = 0\) and \(P_2 = 0\) can be written in the form \(P_1 + \lambda P_2 = 0\), where \(\lambda\) is a scalar parameter. This represents a family of planes, and we can find the specific plane that passes through a given point by solving for \(\lambda\).
Step 2: Key Formula or Approach:
1. Let the two planes be \(P_1 \equiv 2x-y-4=0\) and \(P_2 \equiv y+2z-4=0\).
2. The equation of the required plane is \((2x-y-4) + \lambda(y+2z-4) = 0\).
3. Since this plane passes through the point (1,1,0), substitute \(x=1, y=1, z=0\) into the equation to find \(\lambda\).
4. Substitute the value of \(\lambda\) back into the family of planes equation to get the final equation.
Step 3: Detailed Explanation:
The equation of the plane passing through the intersection of the given planes is:
\[ (2x-y-4) + \lambda(y+2z-4) = 0 \]
This plane passes through the point (1,1,0). So, we substitute these coordinates into the equation:
\[ (2(1) - (1) - 4) + \lambda((1) + 2(0) - 4) = 0 \]
\[ (2 - 1 - 4) + \lambda(1 - 4) = 0 \]
\[ (-3) + \lambda(-3) = 0 \]
\[ -3\lambda = 3 \]
\[ \lambda = -1 \]
Now, substitute \(\lambda = -1\) back into the equation of the family of planes:
\[ (2x-y-4) + (-1)(y+2z-4) = 0 \]
\[ 2x-y-4 - y - 2z + 4 = 0 \]
\[ 2x - 2y - 2z = 0 \]
Divide the entire equation by 2 to simplify:
\[ x - y - z = 0 \]
Step 4: Final Answer:
The equation of the required plane is \(x-y-z=0\), which corresponds to option (C).