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

Options

A. (1, 2, 3)
B. (3, 2, 1)
C. (2, 3, 1)
D. (3, 1, 2)

Answer

To find the equation of the plane passing through the intersection of the planes \( P_1: x + 2y + 3z + 4 = 0 \) and \( P_2: 4x + 3y + 2z + 1 = 0 \), and also through the origin, we can follow these steps:

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

### Step 2: Use the formula for the plane through the intersection
The equation of the plane passing through the intersection of two planes can be expressed as:
\[ P_1 + \lambda P_2 = 0 \]
where \( \lambda \) is a parameter.

Substituting the plane equations:
\[ (x + 2y + 3z + 4) + \lambda(4x + 3y + 2z + 1) = 0 \]

### Step 3: Expand the equation
Expanding the equation gives:
\[ x + 2y + 3z + 4 + \lambda(4x + 3y + 2z + 1) = 0 \]
This simplifies to:
\[ x + 2y + 3z + 4 + 4\lambda x + 3\lambda y + 2\lambda z + \lambda = 0 \]

### Step 4: Combine like terms
Combining the coefficients of \( x, y, z \) and the constant term results in:
\[ (1 + 4\lambda)x + (2 + 3\lambda)y + (3 + 2\lambda)z + (4 + \lambda) = 0 \]

### Step 5: Substitute the origin (0, 0, 0)
Since the plane passes through the origin, we substitute \( (0, 0, 0) \) into the equation:
\[ 4 + \lambda = 0 \]
This gives:
\[ \lambda = -4 \]

### Step 6: Substitute \( \lambda \) back into the equation
Now substitute \( \lambda = -4 \) back into the combined equation:
\[ (1 + 4(-4))x + (2 + 3(-4))y + (3 + 2(-4))z + (4 - 4) = 0 \]
This simplifies to:
\[ (1 - 16)x + (2 - 12)y + (3 - 8)z + 0 = 0 \]
Thus:
\[ -15x - 10y - 5z = 0 \]

### Step 7: Simplify the equation
Dividing the entire equation by -5 gives:
\[ 3x + 2y + z = 0 \]

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

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