The equation of the plane through the line `x+y+z+3=0=2x-y+3z+1` and parallel to the line `x/1=y/2=z/3` is

Options

A. `x-5y+3z=7`
B. `x-5y+3z=-7`
C. `x+5y+z=7`
D. `x+5y+3z=-7`

Explanation

To find the equation of the plane that passes through the line of intersection of the two given planes and is parallel to a specified line, we can follow these steps:

### Step 1: Write the equations of the given planes
The equations of the two planes are:
1. \( P_1: x + y + z + 3 = 0 \)
2. \( P_2: 2x - y + 3z + 1 = 0 \)

### Step 2: Form the family of planes
The equation of a family of planes that passes through the line of intersection of the two planes can be expressed as:
\[ P_1 + \lambda P_2 = 0 \]
Substituting the equations of the planes:
\[ (x + y + z + 3) + \lambda (2x - y + 3z + 1) = 0 \]

### Step 3: Simplify the equation
Expanding and simplifying the equation gives:
\[ x + y + z + 3 + \lambda(2x - y + 3z + 1) = 0 \]
This can be rearranged as:
\[ (1 + 2\lambda)x + (1 - \lambda)y + (1 + 3\lambda)z + (3 + \lambda) = 0 \]

### Step 4: Identify the direction ratios of the given line
The line given by \( \frac{x}{1} = \frac{y}{2} = \frac{z}{3} \) has direction ratios:
\[ \mathbf{b} = (1, 2, 3) \]

### Step 5: Find the normal vector of the plane
The normal vector \( \mathbf{n} \) of the plane can be identified from the coefficients of \( x, y, z \) in the plane equation:
\[ \mathbf{n} = (1 + 2\lambda, 1 - \lambda, 1 + 3\lambda) \]

### Step 6: Set up the perpendicularity condition
Since the plane is parallel to the line, the normal vector \( \mathbf{n} \) must be perpendicular to the direction vector \( \mathbf{b} \). Therefore, we set up the dot product:
\[ \mathbf{n} \cdot \mathbf{b} = 0 \]
This gives:
\[ (1 + 2\lambda) \cdot 1 + (1 - \lambda) \cdot 2 + (1 + 3\lambda) \cdot 3 = 0 \]

### Step 7: Solve for \( \lambda \)
Expanding the dot product:
\[ 1 + 2\lambda + 2 - 2\lambda + 3 + 9\lambda = 0 \]
Combining like terms:
\[ 1 + 2 + 3 + (2\lambda - 2\lambda + 9\lambda) = 0 \]
This simplifies to:
\[ 6 + 9\lambda = 0 \]
Solving for \( \lambda \):
\[ 9\lambda = -6 \implies \lambda = -\frac{2}{3} \]

### Step 8: Substitute \( \lambda \) back into the plane equation
Substituting \( \lambda = -\frac{2}{3} \) back into the equation of the plane:
\[ (1 + 2(-\frac{2}{3}))x + (1 - (-\frac{2}{3}))y + (1 + 3(-\frac{2}{3}))z + (3 - \frac{2}{3}) = 0 \]
This simplifies to:
\[ (1 - \frac{4}{3})x + (1 + \frac{2}{3})y + (1 - 2)z + \frac{7}{3} = 0 \]
Which becomes:
\[ -\frac{1}{3}x + \frac{5}{3}y - z + \frac{7}{3} = 0 \]
Multiplying through by 3 to eliminate fractions:
\[ -x + 5y - 3z + 7 = 0 \implies x - 5y + 3z = 7 \]

### Final Answer
The equation of the required plane is:
\[ x - 5y + 3z = 7 \]