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 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: Form the equation of the plane through the intersection
The equation of a plane that passes through the line of intersection of two planes \( P_1 \) and \( P_2 \) can be expressed as:
\[ P_3: P_1 + \lambda P_2 = 0 \]
Substituting the equations of \( P_1 \) and \( P_2 \):
\[ (x + y + 2z - 4) + \lambda(2x - y + 3z + 5) = 0 \]
Expanding this gives:
\[ x + y + 2z - 4 + 2\lambda x - \lambda y + 3\lambda z + 5\lambda = 0 \]
Combining like terms:
\[ (1 + 2\lambda)x + (1 - \lambda)y + (2 + 3\lambda)z + (-4 + 5\lambda) = 0 \]

### Step 3: Find the direction vector of line segment \( AB \)
Points \( A(1, 3, 4) \) and \( B(3, 2, 7) \) give us the direction vector \( \overrightarrow{AB} \):
\[ \overrightarrow{AB} = B - A = (3 - 1, 2 - 3, 7 - 4) = (2, -1, 3) \]

### Step 4: Find the normal vector of the plane \( P_3 \)
The normal vector of the plane \( P_3 \) can be represented as:
\[ \mathbf{n} = (1 + 2\lambda, 1 - \lambda, 2 + 3\lambda) \]
Since \( P_3 \) is parallel to \( AB \), the dot product of \( \mathbf{n} \) and \( \overrightarrow{AB} \) must equal zero:
\[ \mathbf{n} \cdot \overrightarrow{AB} = 0 \]
This gives us:
\[ (1 + 2\lambda) \cdot 2 + (1 - \lambda)(-1) + (2 + 3\lambda) \cdot 3 = 0 \]
Expanding this:
\[ 2(1 + 2\lambda) - (1 - \lambda) + 3(2 + 3\lambda) = 0 \]
Simplifying:
\[ 2 + 4\lambda - 1 + \lambda + 6 + 9\lambda = 0 \]
Combining like terms:
\[ (4\lambda + \lambda + 9\lambda) + (2 - 1 + 6) = 0 \implies 14\lambda + 7 = 0 \]
Solving for \( \lambda \):
\[ \lambda = -\frac{1}{2} \]

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

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