Find the equation of a plane containing the line of intersection of the planes `x+y+z-6=0a n d2x+3y+4z+5=0` passing through `(1,1,1)` .

Options

A. `20x+23y+26z-69=0`
B. `20x+26y+23z-69=0`
C. `x+y+z-3=0`
D. `2x+3y+4z-9=0`

Explanation

To find the equation of a plane containing the line of intersection of the planes \( P_1: x + y + z - 6 = 0 \) and \( P_2: 2x + 3y + 4z + 5 = 0 \), and passing through the point \( (1, 1, 1) \), we can follow these steps:

### Step 1: Write the equations of the planes
We have two planes given by:
1. \( P_1: x + y + z - 6 = 0 \)
2. \( P_2: 2x + 3y + 4z + 5 = 0 \)

### Step 2: Form the general equation of the plane
The equation of a plane that contains the line of intersection of two planes can be expressed as:
\[ P_1 + \lambda P_2 = 0 \]
where \( \lambda \) is a parameter. Substituting the equations of the planes, we have:
\[ (x + y + z - 6) + \lambda(2x + 3y + 4z + 5) = 0 \]

### Step 3: Expand the equation
Expanding the equation gives:
\[ x + y + z - 6 + \lambda(2x + 3y + 4z + 5) = 0 \]
This can be rewritten as:
\[ (1 + 2\lambda)x + (1 + 3\lambda)y + (1 + 4\lambda)z - (6 + 5\lambda) = 0 \]

### Step 4: Substitute the point (1, 1, 1)
Since the plane passes through the point \( (1, 1, 1) \), we substitute \( x = 1 \), \( y = 1 \), and \( z = 1 \) into the equation:
\[ (1 + 2\lambda)(1) + (1 + 3\lambda)(1) + (1 + 4\lambda)(1) - (6 + 5\lambda) = 0 \]
This simplifies to:
\[ (1 + 2\lambda) + (1 + 3\lambda) + (1 + 4\lambda) - (6 + 5\lambda) = 0 \]
Combining like terms:
\[ 3 + 9\lambda - 6 - 5\lambda = 0 \]
This simplifies to:
\[ -3 + 4\lambda = 0 \]

### Step 5: Solve for \( \lambda \)
Solving for \( \lambda \):
\[ 4\lambda = 3 \quad \Rightarrow \quad \lambda = \frac{3}{4} \]

### Step 6: Substitute \( \lambda \) back into the plane equation
Now we substitute \( \lambda = \frac{3}{4} \) back into the equation of the plane:
\[ (1 + 2 \cdot \frac{3}{4})x + (1 + 3 \cdot \frac{3}{4})y + (1 + 4 \cdot \frac{3}{4})z - (6 + 5 \cdot \frac{3}{4}) = 0 \]
Calculating each term:
- For \( x \): \( 1 + \frac{3}{2} = \frac{5}{2} \)
- For \( y \): \( 1 + \frac{9}{4} = \frac{13}{4} \)
- For \( z \): \( 1 + 3 = 4 \)
- For the constant term: \( 6 + \frac{15}{4} = \frac{39}{4} \)

Thus, the equation becomes:
\[ \frac{5}{2}x + \frac{13}{4}y + 4z - \frac{39}{4} = 0 \]

### Step 7: Eliminate fractions
To eliminate fractions, multiply through by 4:
\[ 10x + 13y + 16z - 39 = 0 \]

### Final Equation of the Plane
The final equation of the plane is:
\[ 10x + 13y + 16z - 39 = 0 \]