Conisder the plane passing through the points A(2,2,1),B(3,4,2) and C(7,0,6).
Which one of the following points lines on the plane ?

Options

A. (1,0,0)
B. (1,0,1)
C. (0,0,1)
D. None of these

Explanation

To determine which point lies on the plane defined by the points A(2,2,1), B(3,4,2), and C(7,0,6), we can follow these steps:

### Step 1: Find the equation of the plane

We can use the determinant method to find the equation of the plane passing through the three points A, B, and C. The general formula for the equation of a plane through three points \( A(x_1, y_1, z_1) \), \( B(x_2, y_2, z_2) \), and \( C(x_3, y_3, z_3) \) is given by:

\[ \begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ x_3 - x_1 & y_3 - y_1 & z_3 - z_1 \end{vmatrix} = 0 \]

Substituting the coordinates of points A, B, and C:

- \( A(2, 2, 1) \) → \( (x_1, y_1, z_1) = (2, 2, 1) \)
- \( B(3, 4, 2) \) → \( (x_2, y_2, z_2) = (3, 4, 2) \)
- \( C(7, 0, 6) \) → \( (x_3, y_3, z_3) = (7, 0, 6) \)

### Step 2: Set up the determinant

The determinant becomes:

\[ \begin{vmatrix} x - 2 & y - 2 & z - 1 \\ 3 - 2 & 4 - 2 & 2 - 1 \\ 7 - 2 & 0 - 2 & 6 - 1 \end{vmatrix} = 0 \]

This simplifies to:

\[ \begin{vmatrix} x - 2 & y - 2 & z - 1 \\ 1 & 2 & 1 \\ 5 & -2 & 5 \end{vmatrix} = 0 \]

### Step 3: Calculate the determinant

Calculating the determinant:

\[ = (x - 2) \begin{vmatrix} 2 & 1 \\ -2 & 5 \end{vmatrix} - (y - 2) \begin{vmatrix} 1 & 1 \\ 5 & 5 \end{vmatrix} + (z - 1) \begin{vmatrix} 1 & 2 \\ 5 & -2 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \( \begin{vmatrix} 2 & 1 \\ -2 & 5 \end{vmatrix} = (2)(5) - (1)(-2) = 10 + 2 = 12 \)
2. \( \begin{vmatrix} 1 & 1 \\ 5 & 5 \end{vmatrix} = (1)(5) - (1)(5) = 5 - 5 = 0 \)
3. \( \begin{vmatrix} 1 & 2 \\ 5 & -2 \end{vmatrix} = (1)(-2) - (2)(5) = -2 - 10 = -12 \)

Substituting these back into the determinant:

\[ = (x - 2)(12) - (y - 2)(0) + (z - 1)(-12) = 0 \]

This simplifies to:

\[ 12(x - 2) - 12(z - 1) = 0 \]

### Step 4: Simplify the equation

Rearranging gives:

\[ 12x - 24 - 12z + 12 = 0 \]

\[ 12x - 12z - 12 = 0 \]

Dividing through by 12:

\[ x - z - 1 = 0 \]

Thus, the equation of the plane is:

\[ x - z = 1 \]

### Step 5: Check which points lie on the plane

Now we need to check which of the given points satisfies the equation \( x - z = 1 \).

1. For point \( P_1(1, 0, 0) \):
\[ 1 - 0 = 1 \quad \text{(satisfies the equation)} \]

2. For point \( P_2(1, 0, 1) \):
\[ 1 - 1 = 0 \quad \text{(does not satisfy the equation)} \]

3. For point \( P_3(0, 0, 1) \):
\[ 0 - 1 = -1 \quad \text{(does not satisfy the equation)} \]

### Conclusion

The point that lies on the plane is \( P_1(1, 0, 0) \).