The equation of the plane passing through the points (1,2,3), (-1,4,2) and (3,1,1) is

Options

A. `5x+y+12z-23=0`
B. `5x+6y+2z-23=0`
C. `x+6y+2z-13=0`
D. `x+y+z-13=0`

Explanation

To find the equation of the plane passing through the points (1, 2, 3), (-1, 4, 2), and (3, 1, 1), we can follow these steps:

### Step 1: Set up the general equation of the plane
The general equation of a plane can be written as:
\[ ax + by + cz = d \]
where \( a, b, c, d \) are constants that we need to determine.

### Step 2: Substitute the first point into the equation
Substituting the point (1, 2, 3) into the equation:
\[ a(1) + b(2) + c(3) = d \]
This simplifies to:
\[ a + 2b + 3c = d \]
Let’s call this **Equation 1**.

### Step 3: Substitute the second point into the equation
Substituting the point (-1, 4, 2) into the equation:
\[ a(-1) + b(4) + c(2) = d \]
This simplifies to:
\[ -a + 4b + 2c = d \]
Let’s call this **Equation 2**.

### Step 4: Substitute the third point into the equation
Substituting the point (3, 1, 1) into the equation:
\[ a(3) + b(1) + c(1) = d \]
This simplifies to:
\[ 3a + b + c = d \]
Let’s call this **Equation 3**.

### Step 5: Set up a system of equations
Now we have three equations:
1. \( a + 2b + 3c = d \) (Equation 1)
2. \( -a + 4b + 2c = d \) (Equation 2)
3. \( 3a + b + c = d \) (Equation 3)

### Step 6: Eliminate \( d \) from the equations
Subtract Equation 2 from Equation 1:
\[ (a + 2b + 3c) - (-a + 4b + 2c) = 0 \]
This simplifies to:
\[ 2a - 2b + c = 0 \quad \text{(Equation 4)} \]

Now, subtract Equation 3 from Equation 1:
\[ (a + 2b + 3c) - (3a + b + c) = 0 \]
This simplifies to:
\[ -2a + b + 2c = 0 \quad \text{(Equation 5)} \]

### Step 7: Solve the system of equations
Now we have two new equations:
1. \( 2a - 2b + c = 0 \) (Equation 4)
2. \( -2a + b + 2c = 0 \) (Equation 5)

From Equation 4, we can express \( c \):
\[ c = 2b - 2a \quad \text{(Equation 6)} \]

Substituting Equation 6 into Equation 5:
\[ -2a + b + 2(2b - 2a) = 0 \]
This simplifies to:
\[ -2a + b + 4b - 4a = 0 \]
Combining like terms gives:
\[ -6a + 5b = 0 \quad \Rightarrow \quad b = \frac{6}{5}a \quad \text{(Equation 7)} \]

### Step 8: Substitute back to find \( a, b, c \)
Substituting Equation 7 back into Equation 6:
\[ c = 2\left(\frac{6}{5}a\right) - 2a = \frac{12}{5}a - 2a = \frac{12}{5}a - \frac{10}{5}a = \frac{2}{5}a \]

### Step 9: Substitute \( a, b, c \) into the original plane equation
Now we have:
- \( b = \frac{6}{5}a \)
- \( c = \frac{2}{5}a \)

Substituting these into the plane equation \( ax + by + cz = d \):
\[ a x + \frac{6}{5}a y + \frac{2}{5}a z = d \]
Factoring out \( a \):
\[ a\left(x + \frac{6}{5}y + \frac{2}{5}z\right) = d \]

### Step 10: Find \( d \) using one of the original points
Using Equation 1 to find \( d \):
\[ d = a + 2\left(\frac{6}{5}a\right) + 3\left(\frac{2}{5}a\right) \]
This simplifies to:
\[ d = a + \frac{12}{5}a + \frac{6}{5}a = a + \frac{18}{5}a = \frac{23}{5}a \]

### Step 11: Final equation of the plane
Substituting \( d \) back into the plane equation:
\[ a\left(x + \frac{6}{5}y + \frac{2}{5}z\right) = \frac{23}{5}a \]
Dividing both sides by \( a \) (assuming \( a \neq 0 \)):
\[ x + \frac{6}{5}y + \frac{2}{5}z = \frac{23}{5} \]
Multiplying through by 5 to eliminate the fractions:
\[ 5x + 6y + 2z = 23 \]

### Final Answer
The equation of the plane is:
\[ \boxed{5x + 6y + 2z = 23} \]