The equation of the plane passing through the line of intersection of the planes `x+y+z=6` and `2x+3y+4z+5=0` and perpendicular to the plane `4x+5y-3z=8` is

Options

A. `x+7y+13z-96=0`
B. `x+7y+13z+96=0`
C. `x+7y-13z-96=0`
D. `x-7y+13z+96=0`

Explanation

To find the equation of the plane that passes through the line of intersection of the planes \(x + y + z = 6\) and \(2x + 3y + 4z + 5 = 0\), and is perpendicular to the plane \(4x + 5y - 3z = 8\), we can follow these steps:

### Step 1: Identify the equations of the given planes
The equations of the two planes are:
1. Plane 1: \(x + y + z = 6\)
2. Plane 2: \(2x + 3y + 4z + 5 = 0\)

### Step 2: Write the general equation of the plane
The equation of a plane that passes through the line of intersection of the two given planes can be expressed as:
\[ \lambda_1 (x + y + z - 6) + \lambda_2 (2x + 3y + 4z + 5) = 0 \]
where \(\lambda_1\) and \(\lambda_2\) are constants.

### Step 3: Simplify the equation
We can rewrite the equation as:
\[ \lambda_1 (x + y + z) + \lambda_2 (2x + 3y + 4z) = 6\lambda_1 - 5\lambda_2 \]
This gives us:
\[ (\lambda_1 + 2\lambda_2)x + (\lambda_1 + 3\lambda_2)y + (\lambda_1 + 4\lambda_2)z = 6\lambda_1 - 5\lambda_2 \]

### Step 4: Use the condition of perpendicularity
For the plane to be perpendicular to the plane \(4x + 5y - 3z = 8\), the normal vectors must satisfy the dot product condition:
\[ (4, 5, -3) \cdot (\lambda_1 + 2\lambda_2, \lambda_1 + 3\lambda_2, \lambda_1 + 4\lambda_2) = 0 \]
This leads to the equation:
\[ 4(\lambda_1 + 2\lambda_2) + 5(\lambda_1 + 3\lambda_2) - 3(\lambda_1 + 4\lambda_2) = 0 \]

### Step 5: Expand and simplify the equation
Expanding the dot product gives:
\[ 4\lambda_1 + 8\lambda_2 + 5\lambda_1 + 15\lambda_2 - 3\lambda_1 - 12\lambda_2 = 0 \]
Combining like terms results in:
\[ (4 + 5 - 3)\lambda_1 + (8 + 15 - 12)\lambda_2 = 0 \]
This simplifies to:
\[ 6\lambda_1 + 11\lambda_2 = 0 \]
From this, we can express \(\lambda_1\) in terms of \(\lambda_2\):
\[ \lambda_1 = -\frac{11}{6}\lambda_2 \]

### Step 6: Substitute back into the plane equation
Substituting \(\lambda_1\) back into the plane equation gives:
\[ -\frac{11}{6}\lambda_2 + 2\lambda_2)x + \left(-\frac{11}{6}\lambda_2 + 3\lambda_2\right)y + \left(-\frac{11}{6}\lambda_2 + 4\lambda_2\right)z = 6\left(-\frac{11}{6}\lambda_2\right) - 5\lambda_2 \]

### Step 7: Simplify the coefficients
This results in:
\[ \left(-\frac{11}{6} + 2\right)\lambda_2 x + \left(-\frac{11}{6} + 3\right)\lambda_2 y + \left(-\frac{11}{6} + 4\right)\lambda_2 z = -11\lambda_2 - 5\lambda_2 \]
Calculating the coefficients:
\[ \left(\frac{1}{6}\lambda_2\right)x + \left(\frac{7}{6}\lambda_2\right)y + \left(\frac{13}{6}\lambda_2\right)z = -16\lambda_2 \]

### Step 8: Divide through by \(\lambda_2\) and simplify
Assuming \(\lambda_2 \neq 0\), we can divide through by \(\lambda_2\):
\[ \frac{1}{6}x + \frac{7}{6}y + \frac{13}{6}z = -16 \]
Multiplying through by 6 gives:
\[ x + 7y + 13z + 96 = 0 \]

### Final Equation
Thus, the equation of the required plane is:
\[ x + 7y + 13z + 96 = 0 \]