The equation of plane through the line of intersection of the planes `2x+3y+4z-7=0,x+y+z-1=0` and perpendicular to the plane `x+y+z-1=0`

Options

A. `x+2y+3z=6`
B. `x-2y+z=6`
C. `2x-y+z=5`
D. `x-2y+6z=3`

Answer

To find the equation of the plane that passes through the line of intersection of the planes \(2x + 3y + 4z - 7 = 0\) and \(x + y + z - 1 = 0\), and is perpendicular to the plane \(x + y + z - 1 = 0\), we can follow these steps:

### Step 1: Identify the given planes
We have two planes:
- Plane 1: \(P_1: 2x + 3y + 4z - 7 = 0\)
- Plane 2: \(P_2: x + y + z - 1 = 0\)

### Step 2: Write the equation of the desired plane
The equation of a plane that passes through the line of intersection of two planes can be expressed as:
\[ P_1 + \lambda P_2 = 0 \]
Substituting the equations of the planes, we get:
\[ (2x + 3y + 4z - 7) + \lambda (x + y + z - 1) = 0 \]
Expanding this gives:
\[ (2 + \lambda)x + (3 + \lambda)y + (4 + \lambda)z - (7 + \lambda) = 0 \]

### Step 3: Find the condition for perpendicularity
The plane we are looking for is perpendicular to the plane \(P_2: x + y + z - 1 = 0\). The normal vector of \(P_2\) is \((1, 1, 1)\). For two planes to be perpendicular, the dot product of their normal vectors must equal zero.

The normal vector of our desired plane is \((2 + \lambda, 3 + \lambda, 4 + \lambda)\). Therefore, we need:
\[ (2 + \lambda) + (3 + \lambda) + (4 + \lambda) = 0 \]
Simplifying this gives:
\[ 2 + \lambda + 3 + \lambda + 4 + \lambda = 0 \]
\[ 9 + 3\lambda = 0 \]
\[ 3\lambda = -9 \implies \lambda = -3 \]

### Step 4: Substitute \(\lambda\) back into the plane equation
Now we substitute \(\lambda = -3\) back into the equation of the plane:
\[ (2 - 3)x + (3 - 3)y + (4 - 3)z - (7 - 3) = 0 \]
This simplifies to:
\[ (-1)x + (0)y + (1)z - 4 = 0 \]
or
\[ -x + z - 4 = 0 \]
Rearranging gives:
\[ x - z + 4 = 0 \]

### Step 5: Final equation of the plane
Thus, the equation of the required plane is:
\[ x - z + 4 = 0 \]