The lines `x + y + z - 3 = 0 = 2x - y + 5z - 6 and x - y - z + 1 = 0 = 2x + 3y + 7z - k` are coplanar then k equals

Options

A. 10
B. 11
C. 12
D. 13

Answer

To find the value of \( k \) such that the given lines are coplanar, we can follow these steps:

### Step 1: Write the equations of the lines
We have two pairs of equations for the lines:
1. \( x + y + z - 3 = 0 \) and \( 2x - y + 5z - 6 = 0 \)
2. \( x - y - z + 1 = 0 \) and \( 2x + 3y + 7z - k = 0 \)

### Step 2: Simplify the first pair of equations
From the first equation:
\[ x + y + z = 3 \quad \text{(1)} \]
From the second equation:
\[ 2x - y + 5z = 6 \quad \text{(2)} \]

### Step 3: Add the two equations from the first pair
Adding equations (1) and (2):
\[ (x + y + z) + (2x - y + 5z) = 3 + 6 \]
This simplifies to:
\[ 3x + 6z = 9 \]
Dividing through by 3 gives:
\[ x + 2z = 3 \quad \text{(3)} \]

### Step 4: Simplify the second pair of equations
From the first equation of the second pair:
\[ x - y - z = -1 \quad \text{(4)} \]
From the second equation:
\[ 2x + 3y + 7z = k \quad \text{(5)} \]

### Step 5: Multiply equation (4) to match coefficients
Multiply equation (4) by 3:
\[ 3x - 3y - 3z = -3 \quad \text{(6)} \]

### Step 6: Add equations (5) and (6)
Now, we add equations (5) and (6):
\[ (2x + 3y + 7z) + (3x - 3y - 3z) = k - 3 \]
This simplifies to:
\[ 5x + 4z = k - 3 \quad \text{(7)} \]

### Step 7: Substitute \( x \) from equation (3) into equation (7)
From equation (3), we can express \( x \) in terms of \( z \):
\[ x = 3 - 2z \]
Substituting this into equation (7):
\[ 5(3 - 2z) + 4z = k - 3 \]
Expanding gives:
\[ 15 - 10z + 4z = k - 3 \]
Combining like terms results in:
\[ 15 - 6z = k - 3 \]

### Step 8: Solve for \( k \)
Rearranging gives:
\[ k = 15 + 3 - 6z \]
Thus:
\[ k = 18 - 6z \quad \text{(8)} \]

### Step 9: Find \( z \) using equation (3)
From equation (3):
\[ x + 2z = 3 \]
Substituting \( x = 1 \) (which we will find later) gives:
\[ 1 + 2z = 3 \implies 2z = 2 \implies z = 1 \]

### Step 10: Substitute \( z \) back into equation (8)
Substituting \( z = 1 \) into equation (8):
\[ k = 18 - 6(1) = 18 - 6 = 12 \]

### Conclusion
Thus, the value of \( k \) for which the lines are coplanar is:
\[ \boxed{12} \]