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

Options

A. `x-5y+3z+5=0`
B. `x-5y+3z=5`
C. `x+5y-3z=5`
D. `x+5y-3=5`

Answer

To find the equation of the plane passing through the line of intersection of the planes \( P_1: x + y + z + 3 = 0 \) and \( P_2: 2x - y + 3z + 2 = 0 \), and parallel to the line given by \( \frac{x}{1} = \frac{y}{2} = \frac{z}{3} \), we can follow these steps:

### Step 1: Identify the equations of the planes
We have two planes:
1. \( P_1: x + y + z + 3 = 0 \)
2. \( P_2: 2x - y + 3z + 2 = 0 \)

### Step 2: Find the equation of the plane through the line of intersection
The equation of a plane through the line of intersection of two planes can be expressed as:
\[ P_1 + \lambda P_2 = 0 \]
Substituting the equations of the planes:
\[ (x + y + z + 3) + \lambda(2x - y + 3z + 2) = 0 \]
Expanding this gives:
\[ x + y + z + 3 + 2\lambda x - \lambda y + 3\lambda z + 2\lambda = 0 \]
Combining like terms:
\[ (1 + 2\lambda)x + (1 - \lambda)y + (1 + 3\lambda)z + (3 + 2\lambda) = 0 \]

### Step 3: Determine the direction ratios of the line
The line given by \( \frac{x}{1} = \frac{y}{2} = \frac{z}{3} \) has direction ratios \( (1, 2, 3) \).

### Step 4: Find the normal vector of the plane
The normal vector of the plane can be derived from the coefficients of \( x, y, z \):
\[ \vec{n} = (1 + 2\lambda, 1 - \lambda, 1 + 3\lambda) \]

### Step 5: Set up the condition for parallelism
For the plane to be parallel to the line, the normal vector of the plane must be perpendicular to the direction ratios of the line. Therefore, their dot product must equal zero:
\[ (1 + 2\lambda) \cdot 1 + (1 - \lambda) \cdot 2 + (1 + 3\lambda) \cdot 3 = 0 \]
Expanding this:
\[ 1 + 2\lambda + 2 - 2\lambda + 3 + 9\lambda = 0 \]
Combining terms:
\[ 6 + 9\lambda = 0 \]
Solving for \( \lambda \):
\[ 9\lambda = -6 \implies \lambda = -\frac{2}{3} \]

### Step 6: Substitute \( \lambda \) back into the plane equation
Substituting \( \lambda = -\frac{2}{3} \) into the equation of the plane:
\[ (1 + 2(-\frac{2}{3}))x + (1 - (-\frac{2}{3}))y + (1 + 3(-\frac{2}{3}))z + (3 + 2(-\frac{2}{3})) = 0 \]
Calculating each term:
- For \( x \): \( 1 - \frac{4}{3} = -\frac{1}{3} \)
- For \( y \): \( 1 + \frac{2}{3} = \frac{5}{3} \)
- For \( z \): \( 1 - 2 = -1 \)
- Constant term: \( 3 - \frac{4}{3} = \frac{5}{3} \)

Thus, the equation becomes:
\[ -\frac{1}{3}x + \frac{5}{3}y - z + \frac{5}{3} = 0 \]
Multiplying through by 3 to eliminate fractions:
\[ -x + 5y - 3z + 5 = 0 \]
Rearranging gives:
\[ x - 5y + 3z = 5 \]

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