The equation of the plane passing through the intersection of `x + 2y + 3z + 4 = 0 and 4x + 3y + 2z+ 1 = 0` and the origin `(0.0, 0)` is

Options

A. `17x+14y+11z=0`
B. `7x+4y+z=0`
C. `x+14y+11z=0`
D. `17x+y+z=0`

Explanation

To find the equation of the plane passing through the intersection of the two given planes and the origin, we can follow these steps:

### Step 1: Write the equations of the given planes.
The equations of the two planes are:
1. \( P_1: x + 2y + 3z + 4 = 0 \)
2. \( P_2: 4x + 3y + 2z + 1 = 0 \)

### Step 2: Form the equation of the required plane.
The required plane can be expressed as a linear combination of the two given planes. Therefore, we can write the equation of the required plane as:
\[ P: (x + 2y + 3z + 4) + \lambda(4x + 3y + 2z + 1) = 0 \]
where \( \lambda \) is a parameter.

### Step 3: Substitute the origin into the equation.
Since the required plane passes through the origin \((0, 0, 0)\), we substitute \( x = 0 \), \( y = 0 \), and \( z = 0 \) into the equation:
\[ (0 + 0 + 0 + 4) + \lambda(0 + 0 + 0 + 1) = 0 \]
This simplifies to:
\[ 4 + \lambda = 0 \]

### Step 4: Solve for \( \lambda \).
From the equation \( 4 + \lambda = 0 \), we find:
\[ \lambda = -4 \]

### Step 5: Substitute \( \lambda \) back into the plane equation.
Now we substitute \( \lambda = -4 \) back into the equation of the required plane:
\[ P: (x + 2y + 3z + 4) - 4(4x + 3y + 2z + 1) = 0 \]
Expanding this gives:
\[ x + 2y + 3z + 4 - 16x - 12y - 8z - 4 = 0 \]
Combining like terms results in:
\[ -15x - 10y - 5z = 0 \]

### Step 6: Simplify the equation.
To simplify, we can factor out \(-5\):
\[ -5(3x + 2y + z) = 0 \]
Thus, the equation of the required plane is:
\[ 3x + 2y + z = 0 \]

### Final Answer:
The equation of the plane passing through the intersection of the two given planes and the origin is:
\[ 3x + 2y + z = 0 \]

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