Let `P_(1):x+y+2z=3 and P_(2):x-2y+z4` be two planes. Let `A(2, 4 5) and B(4, 3, 8)` be two points in space. The equation of plane `P_(3)` through the line of intersection of `P_(1) and P_(2)` such that the length of the projection upon it of the line segment AB is the least, is

Options

A. `2x-y+3z=7`
B. `3y+z+1=0`
C. `x+3y+z+2=0`
D. `3x-3y+4z-11=0`

Answer

To solve the problem step by step, we will find the equation of the plane \( P_3 \) that passes through the line of intersection of the planes \( P_1 \) and \( P_2 \) such that the length of the projection of the line segment \( AB \) onto the plane \( P_3 \) is minimized.

### Step 1: Write the equations of the planes
The equations of the planes are given as:
- \( P_1: x + y + 2z = 3 \)
- \( P_2: x - 2y + z = 4 \)

### Step 2: Form the equation of plane \( P_3 \)
Since the plane \( P_3 \) passes through the line of intersection of planes \( P_1 \) and \( P_2 \), we can express the equation of \( P_3 \) as a linear combination of \( P_1 \) and \( P_2 \):
\[ P_3: P_1 + \lambda P_2 = 0 \]
This gives us:
\[ (x + y + 2z) + \lambda (x - 2y + z) = 3 + 4\lambda \]
Expanding this, we get:
\[ (1 + \lambda)x + (1 - 2\lambda)y + (2 + \lambda)z = 3 + 4\lambda \]

### Step 3: Identify the direction vector of line segment \( AB \)
The points \( A(2, 4, 5) \) and \( B(4, 3, 8) \) give us the direction vector \( \overrightarrow{AB} \):
\[ \overrightarrow{AB} = B - A = (4 - 2, 3 - 4, 8 - 5) = (2, -1, 3) \]

### Step 4: Find the normal vector of plane \( P_3 \)
The normal vector \( \mathbf{n} \) of the plane \( P_3 \) can be represented as:
\[ \mathbf{n} = (1 + \lambda, 1 - 2\lambda, 2 + \lambda) \]

### Step 5: Set up the condition for perpendicularity
For the projection of \( \overrightarrow{AB} \) onto the plane to be minimized, \( \overrightarrow{AB} \) must be perpendicular to the normal vector \( \mathbf{n} \). This gives us the condition:
\[ \overrightarrow{AB} \cdot \mathbf{n} = 0 \]
Calculating this, we have:
\[ (2, -1, 3) \cdot (1 + \lambda, 1 - 2\lambda, 2 + \lambda) = 0 \]
Expanding this:
\[ 2(1 + \lambda) - 1(1 - 2\lambda) + 3(2 + \lambda) = 0 \]
This simplifies to:
\[ 2 + 2\lambda - 1 + 2\lambda + 6 + 3\lambda = 0 \]
Combining like terms:
\[ (2\lambda + 2\lambda + 3\lambda) + (2 - 1 + 6) = 0 \]
\[ 7\lambda + 7 = 0 \]
Thus, we find:
\[ \lambda = -1 \]

### Step 6: Substitute \( \lambda \) back into the equation of \( P_3 \)
Substituting \( \lambda = -1 \) into the equation of \( P_3 \):
\[ P_3: (1 - 1)x + (1 + 2)y + (2 - 1)z = 3 - 4 \]
This simplifies to:
\[ 0x + 3y + z = -1 \]
or:
\[ 3y + z = -1 \]

### Final Equation of Plane \( P_3 \)
The equation of the plane \( P_3 \) is:
\[ 3y + z + 1 = 0 \]