Let `P_(1):x +y+2z-4=0 and P_(2): 2x-y+3z+5=0` be the planes. Let `A(1, 3, 4) and B(3, 2, 7)` be two points in space. The equation of a third plane `P_(3)` through the line of intersection of `P_(1) and P_(2)` and parallel to AB is

Options

A. `x-4y-2z+3=0`
B. `x-4y-2z+9=0`
C. `2x-3y+4z+9=0`
D. `3y+z-13=0`

Answer

To find the equation of the third plane \( P_3 \) that passes through the line of intersection of the planes \( P_1 \) and \( P_2 \) and is parallel to the line segment \( AB \), we can follow these steps:

### Step 1: Write down the equations of the planes
The equations of the planes are given as:
\[ P_1: x + y + 2z - 4 = 0 \]
\[ P_2: 2x - y + 3z + 5 = 0 \]

### Step 2: Find the direction vector of line segment \( AB \)
The points \( A(1, 3, 4) \) and \( B(3, 2, 7) \) can be used to find the direction vector \( \vec{AB} \):
\[ \vec{AB} = B - A = (3 - 1, 2 - 3, 7 - 4) = (2, -1, 3) \]

### Step 3: Form the equation of the plane \( P_3 \)
The equation of the plane \( P_3 \) can be expressed as a linear combination of the equations of planes \( P_1 \) and \( P_2 \):
\[ P_3: P_1 + \lambda P_2 = 0 \]
Substituting the equations of \( P_1 \) and \( P_2 \):
\[ (x + y + 2z - 4) + \lambda(2x - y + 3z + 5) = 0 \]
Expanding this gives:
\[ x + y + 2z - 4 + \lambda(2x - y + 3z + 5) = 0 \]

### Step 4: Combine like terms
Combining the terms, we have:
\[ (1 + 2\lambda)x + (1 - \lambda)y + (2 + 3\lambda)z + (-4 + 5\lambda) = 0 \]
This can be written as:
\[ (1 + 2\lambda)x + (1 - \lambda)y + (2 + 3\lambda)z + (5\lambda - 4) = 0 \]

### Step 5: Find the normal vector of the plane
The normal vector of the plane \( P_3 \) is given by the coefficients of \( x, y, z \):
\[ \vec{n} = (1 + 2\lambda, 1 - \lambda, 2 + 3\lambda) \]

### Step 6: Set the condition for parallelism
For the plane \( P_3 \) to be parallel to the line segment \( AB \), the normal vector \( \vec{n} \) must be perpendicular to the direction vector \( \vec{AB} = (2, -1, 3) \). This gives us the dot product condition:
\[ (1 + 2\lambda) \cdot 2 + (1 - \lambda)(-1) + (2 + 3\lambda) \cdot 3 = 0 \]

### Step 7: Solve for \( \lambda \)
Expanding the dot product:
\[ 2(1 + 2\lambda) - (1 - \lambda) + 3(2 + 3\lambda) = 0 \]
This simplifies to:
\[ 2 + 4\lambda - 1 + \lambda + 6 + 9\lambda = 0 \]
Combining like terms:
\[ (4\lambda + \lambda + 9\lambda) + (2 - 1 + 6) = 0 \]
\[ 14\lambda + 7 = 0 \]
Solving for \( \lambda \):
\[ 14\lambda = -7 \implies \lambda = -\frac{1}{2} \]

### Step 8: Substitute \( \lambda \) back into the equation of \( P_3 \)
Substituting \( \lambda = -\frac{1}{2} \) into the equation of \( P_3 \):
\[ (1 + 2(-\frac{1}{2}))x + (1 - (-\frac{1}{2}))y + (2 + 3(-\frac{1}{2}))z + (5(-\frac{1}{2}) - 4) = 0 \]
This simplifies to:
\[ (1 - 1)x + (1 + \frac{1}{2})y + (2 - \frac{3}{2})z - \frac{5}{2} - 4 = 0 \]
\[ 0x + \frac{3}{2}y + \frac{1}{2}z - \frac{13}{2} = 0 \]
Multiplying through by 2 to eliminate fractions:
\[ 3y + z - 13 = 0 \]

### Final Equation
Thus, the equation of the plane \( P_3 \) is:
\[ 3y + z - 13 = 0 \]