The equation of the plane through the intersection of the planes `x+y+z=1` and `2x+3y-z+4 = 0` and parallel to x-axis is

Options

A. `y-3z+6=0`
B. `3y-z+6=0`
C. `y+3z+6=0`
D. `3y-2z+6=0`

Answer

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

### Step 1: Write the equations of the given planes
The equations of the two planes are:
1. Plane 1: \(x + y + z = 1\)
2. Plane 2: \(2x + 3y - z + 4 = 0\)

### Step 2: Express the equation of the plane through the intersection
The equation of the plane through the intersection of the two planes can be expressed as:
\[ P_1 + \lambda P_2 = 0 \]
Substituting the equations of the planes:
\[ (x + y + z - 1) + \lambda (2x + 3y - z + 4) = 0 \]
Expanding this, we get:
\[ x + y + z - 1 + \lambda (2x + 3y - z + 4) = 0 \]
This simplifies to:
\[ (1 + 2\lambda)x + (1 + 3\lambda)y + (1 - \lambda)z + (-1 + 4\lambda) = 0 \]

### Step 3: Condition for the plane to be parallel to the x-axis
For the plane to be parallel to the x-axis, the coefficient of \(x\) must be zero. Therefore, we set:
\[ 1 + 2\lambda = 0 \]
Solving for \(\lambda\):
\[ 2\lambda = -1 \implies \lambda = -\frac{1}{2} \]

### Step 4: Substitute \(\lambda\) back into the plane equation
Now, substitute \(\lambda = -\frac{1}{2}\) into the equation:
\[ (1 + 2(-\frac{1}{2}))x + (1 + 3(-\frac{1}{2}))y + (1 - (-\frac{1}{2}))z + (-1 + 4(-\frac{1}{2})) = 0 \]
This simplifies to:
\[ 0x + (-\frac{1}{2})y + \frac{3}{2}z - 3 = 0 \]
Rearranging gives:
\[ -\frac{1}{2}y + \frac{3}{2}z = 3 \]
Multiplying through by -2 to eliminate the fractions:
\[ y - 3z = -6 \]

### Final Equation
Thus, the equation of the plane is:
\[ y - 3z = -6 \]