The equation of the plane passing through the intersection of the planes `2x+y+2z=9,4x-5y-4z=1` and the point (3,2,1) is

Options

A. `10x-2y+2z=28`
B. `10x+2y+2z=28`
C. `10x+2y-2z=28`
D. `10x-2y-2z=14`

Explanation

To find the equation of the plane that passes through the intersection of the planes given by the equations \(2x + y + 2z = 9\) and \(4x - 5y - 4z = 1\), and also passes through the point \((3, 2, 1)\), we can follow these steps:

### Step 1: Write the equations of the two planes
The equations of the two planes are:
1. \(2x + y + 2z - 9 = 0\) (Plane 1)
2. \(4x - 5y - 4z - 1 = 0\) (Plane 2)

### Step 2: Form the equation of the required plane
The equation of the plane that passes through the intersection of the two given planes can be expressed in the form:
\[ (2x + y + 2z - 9) + \lambda(4x - 5y - 4z - 1) = 0 \]
where \(\lambda\) is a parameter.

### Step 3: Substitute the point into the plane equation
We need to find the value of \(\lambda\) such that the plane passes through the point \((3, 2, 1)\). Substitute \(x = 3\), \(y = 2\), and \(z = 1\) into the equation:
\[ (2(3) + 2 + 2(1) - 9) + \lambda(4(3) - 5(2) - 4(1) - 1) = 0 \]

### Step 4: Simplify the equation
Calculating the first part:
\[ 2(3) + 2 + 2(1) - 9 = 6 + 2 + 2 - 9 = 1 \]
Now, calculating the second part:
\[ 4(3) - 5(2) - 4(1) - 1 = 12 - 10 - 4 - 1 = -3 \]
Thus, the equation becomes:
\[ 1 + \lambda(-3) = 0 \]

### Step 5: Solve for \(\lambda\)
Rearranging gives:
\[ -3\lambda = -1 \implies \lambda = \frac{1}{3} \]

### Step 6: Substitute \(\lambda\) back into the plane equation
Now substitute \(\lambda = \frac{1}{3}\) back into the equation of the plane:
\[ (2x + y + 2z - 9) + \frac{1}{3}(4x - 5y - 4z - 1) = 0 \]
Multiply through by 3 to eliminate the fraction:
\[ 3(2x + y + 2z - 9) + (4x - 5y - 4z - 1) = 0 \]

### Step 7: Expand and simplify
Expanding gives:
\[ 6x + 3y + 6z - 27 + 4x - 5y - 4z - 1 = 0 \]
Combine like terms:
\[ (6x + 4x) + (3y - 5y) + (6z - 4z) - 28 = 0 \]
This simplifies to:
\[ 10x - 2y + 2z - 28 = 0 \]

### Step 8: Final equation of the plane
Thus, the equation of the required plane is:
\[ 10x - 2y + 2z = 28 \]