The direction ratios of a normal to the plane passing throuhg (0,0,1) ,(0,1,2) and (1,2,3) are proportional to

Options

A. 0,1,-1
B. 1,0,-1
C. 0,0,-1
D. 1,0,1

Explanation

To find the direction ratios of the normal to the plane passing through the points \( P(0, 0, 1) \), \( Q(0, 1, 2) \), and \( R(1, 2, 3) \), we will follow these steps:

### Step 1: Identify the Points
The points given are:
- \( P(0, 0, 1) \)
- \( Q(0, 1, 2) \)
- \( R(1, 2, 3) \)

### Step 2: Find Vectors in the Plane
We need to find two vectors that lie in the plane formed by these three points. We can do this by calculating the vectors \( \overrightarrow{PQ} \) and \( \overrightarrow{PR} \).

1. **Calculate \( \overrightarrow{PQ} \)**:
\[ \overrightarrow{PQ} = Q - P = (0 - 0, 1 - 0, 2 - 1) = (0, 1, 1) \]

2. **Calculate \( \overrightarrow{PR} \)**:
\[ \overrightarrow{PR} = R - P = (1 - 0, 2 - 0, 3 - 1) = (1, 2, 2) \]

### Step 3: Use the Cross Product to Find the Normal Vector
The normal vector to the plane can be found using the cross product of the two vectors \( \overrightarrow{PQ} \) and \( \overrightarrow{PR} \).

\[ \overrightarrow{N} = \overrightarrow{PQ} \times \overrightarrow{PR} \]

Using the determinant method for the cross product:
\[ \overrightarrow{N} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 1 \\ 1 & 2 & 2 \end{vmatrix} \]

### Step 4: Calculate the Determinant
Calculating the determinant:
\[ \overrightarrow{N} = \hat{i} \begin{vmatrix} 1 & 1 \\ 2 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 0 & 1 \\ 1 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 0 & 1 \\ 1 & 2 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} 1 & 1 \\ 2 & 2 \end{vmatrix} = 1 \cdot 2 - 1 \cdot 2 = 0 \)
2. \( \begin{vmatrix} 0 & 1 \\ 1 & 2 \end{vmatrix} = 0 \cdot 2 - 1 \cdot 1 = -1 \)
3. \( \begin{vmatrix} 0 & 1 \\ 1 & 2 \end{vmatrix} = 0 \cdot 2 - 1 \cdot 1 = -1 \)

Putting it all together:
\[ \overrightarrow{N} = 0 \hat{i} - (-1) \hat{j} + (-1) \hat{k} = 0 \hat{i} + 1 \hat{j} - 1 \hat{k} \]
Thus, \( \overrightarrow{N} = (0, 1, -1) \).

### Step 5: Direction Ratios of the Normal
The direction ratios of the normal vector are \( (0, 1, -1) \).

### Final Answer
The direction ratios of the normal to the plane are proportional to \( (0, 1, -1) \).

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