Conisder the plane passing through the points A(2,2,1),B(3,4,2) and C(7,0,6).
What are the direction ratios of the normal to the plane ?

Options

A. lt 1,0,1 gt
B. lt 0,1,0 gt
C. lt1,0,-1gt
D. None of these

Explanation

To find the direction ratios of the normal to the plane passing through the points A(2, 2, 1), B(3, 4, 2), and C(7, 0, 6), we can follow these steps:

### Step 1: Identify the points
We have three points:
- A(2, 2, 1) → (x1, y1, z1)
- B(3, 4, 2) → (x2, y2, z2)
- C(7, 0, 6) → (x3, y3, z3)

### Step 2: Formulate the equation of the plane
The equation of the plane can be derived using the determinant of a matrix formed by the coordinates of the points. The general form of the equation of a plane passing through three points A, B, and C is given by:

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

### Step 3: Substitute the coordinates into the determinant
Substituting the coordinates of points A, B, and C into the determinant:

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

This simplifies to:

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

### Step 4: Calculate the determinant
Now we calculate the determinant:

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

Calculating each of the 2x2 determinants:

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

2. \(\begin{vmatrix} 1 & 1 \\ 5 & 5 \end{vmatrix} = (1 \cdot 5) - (1 \cdot 5) = 5 - 5 = 0\)

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

Substituting back into the equation gives us:

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

This simplifies to:

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

### Step 5: Rearranging the equation
Rearranging gives:

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

### Step 6: Identify the direction ratios
The coefficients of \(x\), \(y\), and \(z\) in the plane equation give us the direction ratios of the normal to the plane. From the equation \(12x + 0y - 12z - 12 = 0\), we can see that the direction ratios are:

\[ (12, 0, -12) \]

### Final Answer
The direction ratios of the normal to the plane are \(12, 0, -12\).

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