The plane through the intersection of the planes `x+y+z=1` and `2x+3y-z+4=0` and parallel to Y-axis also passes through the point

Options

A. (3,2,1)
B. (3,3,-1)
C. (-3,1,1)
D. (-3,0,1)

Explanation

To solve the problem, we need to find the equation of a plane that passes through the intersection of two given planes and is parallel to the Y-axis. Let's break down the solution step by step.

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

### Step 2: Write 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:
\[ (x + y + z - 1) + \lambda(2x + 3y - z + 4) = 0 \]

### Step 3: Expand the equation
Expanding the equation gives:
\[ x + y + z - 1 + \lambda(2x + 3y - z + 4) = 0 \]
This simplifies to:
\[ (1 + 2\lambda)x + (1 + 3\lambda)y + (1 - \lambda)z + (4\lambda - 1) = 0 \]

### Step 4: Identify coefficients
From the expanded equation, we can identify the coefficients:
- Coefficient of \(x\): \(a = 1 + 2\lambda\)
- Coefficient of \(y\): \(b = 1 + 3\lambda\)
- Coefficient of \(z\): \(c = 1 - \lambda\)
- Constant term: \(d = 4\lambda - 1\)

### Step 5: Condition for the plane to be parallel to the Y-axis
For the plane to be parallel to the Y-axis, the coefficient of \(y\) must not affect the normal vector. Thus, we need:
\[ a_1 \cdot 0 + b_1 \cdot 1 + c_1 \cdot 0 = 0 \]
This simplifies to:
\[ b = 1 + 3\lambda = 0 \]

### Step 6: Solve for \(\lambda\)
Setting \(b = 0\):
\[ 1 + 3\lambda = 0 \implies 3\lambda = -1 \implies \lambda = -\frac{1}{3} \]

### Step 7: Substitute \(\lambda\) back into the plane equation
Substituting \(\lambda = -\frac{1}{3}\) into the coefficients:
- \(a = 1 + 2(-\frac{1}{3}) = 1 - \frac{2}{3} = \frac{1}{3}\)
- \(b = 1 + 3(-\frac{1}{3}) = 1 - 1 = 0\)
- \(c = 1 - (-\frac{1}{3}) = 1 + \frac{1}{3} = \frac{4}{3}\)
- \(d = 4(-\frac{1}{3}) - 1 = -\frac{4}{3} - 1 = -\frac{4}{3} - \frac{3}{3} = -\frac{7}{3}\)

### Step 8: Write the final equation of the plane
The equation of the plane can be written as:
\[ \frac{1}{3}x + 0y + \frac{4}{3}z - \frac{7}{3} = 0 \]
Multiplying through by 3 to eliminate the fractions:
\[ x + 4z - 7 = 0 \]

### Final Answer
The equation of the required plane is:
\[ x + 4z - 7 = 0 \]