The equation of the plane containing the line `2x-5y+z=3, x+y+4z=5` and parallel to the plane `x+3y+6z=1`, is

Options

A. `x+ 3y + 6z =7`
B. `2x + 6y+ 12z =-13`
C. `2x + 6y + 12 z=13`
D. `x+ 3y+ 6z=-7`

Answer

To find the equation of the plane containing the line defined by the intersection of the two planes \(2x - 5y + z = 3\) and \(x + y + 4z = 5\), and which is also parallel to the plane \(x + 3y + 6z = 1\), we can follow these steps:

### Step 1: Write the equations of the given planes
The equations of the two planes are:
1. \(P_1: 2x - 5y + z = 3\)
2. \(P_2: x + y + 4z = 5\)

### Step 2: Find the general equation of the plane through the intersection of the two planes
The general equation of a plane that passes through the intersection of two planes can be expressed as:
\[ P_1 + \lambda P_2 = 0 \]
Substituting the equations of the planes, we have:
\[ (2x - 5y + z - 3) + \lambda (x + y + 4z - 5) = 0 \]
This simplifies to:
\[ (2 + \lambda)x + (-5 + \lambda)y + (1 + 4\lambda)z - (3 + 5\lambda) = 0 \]

### Step 3: Identify the normal vector of the plane
The normal vector \(N\) of the plane we derived is:
\[ N = (2 + \lambda, -5 + \lambda, 1 + 4\lambda) \]

### Step 4: Find the normal vector of the given parallel plane
The normal vector of the plane \(x + 3y + 6z = 1\) is:
\[ N' = (1, 3, 6) \]

### Step 5: Set up the proportionality condition for parallel planes
Since the required plane is parallel to the given plane, their normal vectors must be proportional:
\[ (2 + \lambda, -5 + \lambda, 1 + 4\lambda) = k(1, 3, 6) \]
for some scalar \(k\).

### Step 6: Set up equations based on the proportionality
From the proportionality, we can set up the following equations:
1. \(2 + \lambda = k\)
2. \(-5 + \lambda = 3k\)
3. \(1 + 4\lambda = 6k\)

### Step 7: Solve the equations
From the first equation, we can express \(k\) in terms of \(\lambda\):
\[ k = 2 + \lambda \]

Substituting \(k\) into the second equation:
\[ -5 + \lambda = 3(2 + \lambda) \]
This simplifies to:
\[ -5 + \lambda = 6 + 3\lambda \]
Rearranging gives:
\[ -5 - 6 = 3\lambda - \lambda \implies -11 = 2\lambda \implies \lambda = -\frac{11}{2} \]

### Step 8: Substitute \(\lambda\) back to find the equation of the plane
Now substitute \(\lambda\) back into the equation of the plane:
\[ (2 - \frac{11}{2})x + (-5 - \frac{11}{2})y + (1 + 4(-\frac{11}{2}))z - (3 + 5(-\frac{11}{2})) = 0 \]
This simplifies to:
\[ -\frac{7}{2}x - \frac{21}{2}y - 21z + \frac{49}{2} = 0 \]
Multiplying through by \(-2\) to eliminate fractions gives:
\[ 7x + 21y + 42z - 49 = 0 \]
or
\[ x + 3y + 6z = 7 \]

### Final Answer
The equation of the required plane is:
\[ \boxed{x + 3y + 6z = 7} \]