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

Explanation

To find the equation of the plane through the intersection of the planes \(x + y + z = 1\) and \(2x + 3y - z + 4 = 0\) that is 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: \(P_1: x + y + z - 1 = 0\)
2. Plane 2: \(P_2: 2x + 3y - z + 4 = 0\)

### Step 2: Form the family of planes
The equation of the plane through the intersection of the two planes can be expressed as:
\[ P_3: P_1 + \lambda P_2 = 0 \]
Substituting the equations of \(P_1\) and \(P_2\):
\[ (x + y + z - 1) + \lambda(2x + 3y - z + 4) = 0 \]
Expanding this, we get:
\[ x + y + z - 1 + 2\lambda x + 3\lambda y - \lambda z + 4\lambda = 0 \]
Combining like terms, we have:
\[ (1 + 2\lambda)x + (1 + 3\lambda)y + (1 - \lambda)z + (4\lambda - 1) = 0 \]

### 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 not be zero, and the coefficients of \(y\) and \(z\) must be such that the normal vector of the plane is perpendicular to the direction vector of the x-axis, which is \((1, 0, 0)\).

The normal vector of the plane \(P_3\) is given by:
\[ N = (1 + 2\lambda, 1 + 3\lambda, 1 - \lambda) \]
The condition for \(N\) to be perpendicular to the x-axis direction vector is:
\[ N \cdot (1, 0, 0) = 0 \]
This gives:
\[ 1 + 2\lambda = 0 \]

### Step 4: Solve for \(\lambda\)
Solving for \(\lambda\):
\[ 2\lambda = -1 \implies \lambda = -\frac{1}{2} \]

### Step 5: Substitute \(\lambda\) back into the plane equation
Now substitute \(\lambda = -\frac{1}{2}\) back into the equation of the plane \(P_3\):
\[ (1 + 2(-\frac{1}{2}))x + (1 + 3(-\frac{1}{2}))y + (1 - (-\frac{1}{2}))z + (4(-\frac{1}{2}) - 1) = 0 \]
Simplifying each term:
\[ (1 - 1)x + (1 - \frac{3}{2})y + (1 + \frac{1}{2})z - 2 = 0 \]
This simplifies to:
\[ 0x - \frac{1}{2}y + \frac{3}{2}z - 2 = 0 \]
Multiplying through by -2 to eliminate fractions:
\[ y - 3z + 4 = 0 \]

### Final Equation
Thus, the equation of the required plane is:
\[ y - 3z + 4 = 0 \]