The equation of a plane through the line of intersection of planes `2x+3y+z-1=0 and x+5y-2z+7=0` and parallel to line `y=0=z` :

Options

A. `4x+7y-5z+15=0`
B. `13y-3z+13=0`
C. `7x-5y+15=0`
D. `7y-5z-15=0`

Answer

To find the equation of a plane that passes through the line of intersection of the given planes and is parallel to the line \( y = 0 = z \), we can follow these steps:

### Step 1: Identify the equations of the given planes
The equations of the planes are:
1. \( 2x + 3y + z - 1 = 0 \) (Plane 1)
2. \( x + 5y - 2z + 7 = 0 \) (Plane 2)

### Step 2: Write the equation of the family of planes
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:
\[ (2x + 3y + z - 1) + \lambda (x + 5y - 2z + 7) = 0 \]

### Step 3: Expand the equation
Expanding the equation gives:
\[ 2x + 3y + z - 1 + \lambda x + 5\lambda y - 2\lambda z + 7\lambda = 0 \]
Combining like terms:
\[ (2 + \lambda)x + (3 + 5\lambda)y + (1 - 2\lambda)z + (-1 + 7\lambda) = 0 \]

### Step 4: Identify the normal vector
The normal vector \( \mathbf{n} \) of the plane is given by:
\[ \mathbf{n} = (2 + \lambda, 3 + 5\lambda, 1 - 2\lambda) \]

### Step 5: Determine the condition for parallelism
Since the plane is parallel to the line \( y = 0 = z \), it means the direction vector of the line is \( \mathbf{d} = (1, 0, 0) \). For the plane to be parallel to this line, the normal vector must be perpendicular to the direction vector. Thus, we set the dot product to zero:
\[ \mathbf{n} \cdot \mathbf{d} = (2 + \lambda) \cdot 1 + (3 + 5\lambda) \cdot 0 + (1 - 2\lambda) \cdot 0 = 0 \]
This simplifies to:
\[ 2 + \lambda = 0 \]

### Step 6: Solve for \( \lambda \)
From the equation \( 2 + \lambda = 0 \), we find:
\[ \lambda = -2 \]

### Step 7: Substitute \( \lambda \) back into the plane equation
Now substitute \( \lambda = -2 \) back into the equation of the plane:
\[ (2 - 2)x + (3 - 10)y + (1 + 4)z + (-1 - 14) = 0 \]
This simplifies to:
\[ 0x - 7y + 5z - 15 = 0 \]
Rearranging gives:
\[ 7y - 5z + 15 = 0 \]

### Final Answer
The required equation of the plane is:
\[ 7y - 5z + 15 = 0 \]