The equation of the plane passing through the line of intersection of the planes x + y + 2z = 1,2x + 3y + 4z = 7, and perpendicular to the plane x - 5y + 3z = 5 is given by:

Options

A. A. x + 2y + 3z - 6=0
B. B. x + 2y + 3x + 6 = 0
C. C. 5x +7y +10z - 15=0
D. D. 3x + 4y + 5z +8=0

Answer

To find the equation of the plane that passes through the line of intersection of the planes \( x + y + 2z = 1 \) and \( 2x + 3y + 4z = 7 \), and is perpendicular to the plane \( x - 5y + 3z = 5 \), we can follow these steps:

### Step 1: Write the equation of the required plane
The equation of the required plane can be expressed as a linear combination of the two given planes. Therefore, we can write:
\[ (x + y + 2z - 1) + \lambda(2x + 3y + 4z - 7) = 0 \]
where \( \lambda \) is a scalar.

### Step 2: Expand the equation
Expanding the equation gives:
\[ x + y + 2z - 1 + \lambda(2x + 3y + 4z - 7) = 0 \]
This simplifies to:
\[ x + y + 2z - 1 + 2\lambda x + 3\lambda y + 4\lambda z - 7\lambda = 0 \]

### Step 3: Combine like terms
Combining the like terms, we have:
\[ (1 + 2\lambda)x + (1 + 3\lambda)y + (2 + 4\lambda)z - (1 + 7\lambda) = 0 \]

### Step 4: Identify the coefficients
From the equation, we can identify the coefficients:
- Coefficient of \( x \): \( 1 + 2\lambda \)
- Coefficient of \( y \): \( 1 + 3\lambda \)
- Coefficient of \( z \): \( 2 + 4\lambda \)
- Constant term: \( -(1 + 7\lambda) \)

### Step 5: Use the perpendicularity condition
Since the required plane is perpendicular to the plane \( x - 5y + 3z = 5 \), we can use the condition for perpendicularity:
\[ (1 + 2\lambda)(1) + (1 + 3\lambda)(-5) + (2 + 4\lambda)(3) = 0 \]

### Step 6: Substitute and simplify
Substituting and simplifying gives:
\[ (1 + 2\lambda) - 5(1 + 3\lambda) + 3(2 + 4\lambda) = 0 \]
Expanding this:
\[ 1 + 2\lambda - 5 - 15\lambda + 6 + 12\lambda = 0 \]
Combining like terms:
\[ (1 - 5 + 6) + (2\lambda - 15\lambda + 12\lambda) = 0 \]
This simplifies to:
\[ 2 - 1\lambda = 0 \]
Thus:
\[ \lambda = 2 \]

### Step 7: Substitute \( \lambda \) back into the plane equation
Now substituting \( \lambda = 2 \) back into the equation of the plane:
\[ (1 + 2(2))x + (1 + 3(2))y + (2 + 4(2))z - (1 + 7(2)) = 0 \]
This gives:
\[ (1 + 4)x + (1 + 6)y + (2 + 8)z - (1 + 14) = 0 \]
Thus:
\[ 5x + 7y + 10z - 15 = 0 \]

### Final Answer
The equation of the required plane is:
\[ 5x + 7y + 10z - 15 = 0 \]