Which of the following are equation for the plane passing through the points `P(1,1,-1),Q(3,0,2)and R(-2,1,0)`? (a) `(2hati-3hatj+3hatk).((x+2)hati+(y-1)hatj+zhatk)=0` (b) `x=3-t,y=-11t,z=2-3t` (c) `(x+2)+11(y-1)=3z` (d) `(2hati-hatj+3hatk)xx(-3hati+hatj).((x+2)hati+(y-1)hatj+zhatk)=0`

Options

A. `(2hati-3hatj+3hatk).((x+2)hati+(y-1)hatj+zhatk)=0`
B. `x=3-t,y=-11t,z=2-3t`
C. `(x+2)+11(y-1)=3z`
D. `(2hati-hatj+3hatk)xx(-3hati+hatj).((x+2)hati+(y-1)hatj+zhatk)=0`

Explanation

To determine which of the given options represents the equation of the plane passing through the points \( P(1,1,-1) \), \( Q(3,0,2) \), and \( R(-2,1,0) \), we can follow these steps:

### Step 1: Find the vectors PQ and PR
First, we need to find the vectors \( \overrightarrow{PQ} \) and \( \overrightarrow{PR} \).

\[ \overrightarrow{PQ} = Q - P = (3 - 1, 0 - 1, 2 - (-1)) = (2, -1, 3) \]
\[ \overrightarrow{PR} = R - P = (-2 - 1, 1 - 1, 0 - (-1)) = (-3, 0, 1) \]

### Step 2: Find the normal vector to the plane
Next, we find the normal vector \( \mathbf{n} \) to the plane by taking the cross product of \( \overrightarrow{PQ} \) and \( \overrightarrow{PR} \).

\[ \mathbf{n} = \overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 3 \\ -3 & 0 & 1 \end{vmatrix} \]

Calculating the determinant:

\[ \mathbf{n} = \mathbf{i}((-1)(1) - (3)(0)) - \mathbf{j}((2)(1) - (3)(-3)) + \mathbf{k}((2)(0) - (-1)(-3)) \]
\[ = \mathbf{i}(-1) - \mathbf{j}(2 + 9) + \mathbf{k}(0 - 3) \]
\[ = -\mathbf{i} - 11\mathbf{j} - 3\mathbf{k} \]
Thus, the normal vector is \( \mathbf{n} = (-1, -11, -3) \).

### Step 3: Write the equation of the plane
The equation of a plane can be expressed in the form:
\[ n_1(x - x_0) + n_2(y - y_0) + n_3(z - z_0) = 0 \]
Using point \( P(1, 1, -1) \) and the normal vector \( (-1, -11, -3) \):

\[ -1(x - 1) - 11(y - 1) - 3(z + 1) = 0 \]
Expanding this gives:
\[ -x + 1 - 11y + 11 - 3z - 3 = 0 \]
\[ -x - 11y - 3z + 9 = 0 \]
Rearranging gives:
\[ x + 11y + 3z = 9 \]

### Step 4: Check the options
Now, we need to check which of the given options matches this equation.

- **Option (a)**: \((2\hat{i} - 3\hat{j} + 3\hat{k}) \cdot ((x + 2)\hat{i} + (y - 1)\hat{j} + z\hat{k}) = 0\) - This does not match our derived equation.
- **Option (b)**: \(x = 3 - t, y = -11t, z = 2 - 3t\) - This represents a line, not a plane.
- **Option (c)**: \((x + 2) + 11(y - 1) = 3z\) - Rearranging gives \(x + 11y - 3z + 9 = 0\), which matches our derived equation.
- **Option (d)**: \((2\hat{i} - \hat{j} + 3\hat{k}) \times (-3\hat{i} + \hat{j}) \cdot ((x + 2)\hat{i} + (y - 1)\hat{j} + z\hat{k}) = 0\) - This is a more complex expression but can be checked to see if it simplifies to our plane equation.

### Conclusion
The correct options for the plane passing through points \( P, Q, R \) are:
- **Option (c)**: \( (x + 2) + 11(y - 1) = 3z \)