The unit vector perendicular to the plane determined by `P(1,−1,2),Q(2,0,−1) and R(0,2,1)`.

Explanation

To find the unit vector perpendicular to the plane determined by the points \( P(1, -1, 2) \), \( Q(2, 0, -1) \), and \( R(0, 2, 1) \), we will follow these steps:

### Step 1: Find the vectors \( \vec{PQ} \) and \( \vec{PR} \)

1. **Calculate \( \vec{PQ} \)**:
\[ \vec{PQ} = Q - P = (2 - 1, 0 - (-1), -1 - 2) = (1, 1, -3) \]
So, \( \vec{PQ} = \hat{i} + \hat{j} - 3\hat{k} \).

2. **Calculate \( \vec{PR} \)**:
\[ \vec{PR} = R - P = (0 - 1, 2 - (-1), 1 - 2) = (-1, 3, -1) \]
So, \( \vec{PR} = -\hat{i} + 3\hat{j} - \hat{k} \).

### Step 2: Find the cross product \( \vec{PQ} \times \vec{PR} \)

To find the vector perpendicular to the plane formed by \( P, Q, R \), we need to compute the cross product \( \vec{PQ} \times \vec{PR} \).

\[ \vec{PQ} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -3 \\ -1 & 3 & -1 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i} \begin{vmatrix} 1 & -3 \\ 3 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -3 \\ -1 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ -1 & 3 \end{vmatrix} \]

Calculating each determinant:

1. \( \begin{vmatrix} 1 & -3 \\ 3 & -1 \end{vmatrix} = (1)(-1) - (-3)(3) = -1 + 9 = 8 \)
2. \( \begin{vmatrix} 1 & -3 \\ -1 & -1 \end{vmatrix} = (1)(-1) - (-3)(-1) = -1 - 3 = -4 \)
3. \( \begin{vmatrix} 1 & 1 \\ -1 & 3 \end{vmatrix} = (1)(3) - (1)(-1) = 3 + 1 = 4 \)

Putting it all together:

\[ \vec{PQ} \times \vec{PR} = 8\hat{i} + 4\hat{j} + 4\hat{k} \]

### Step 3: Find the magnitude of the cross product

Now, we need to find the magnitude of the vector \( \vec{PQ} \times \vec{PR} \):

\[ |\vec{PQ} \times \vec{PR}| = \sqrt{8^2 + 4^2 + 4^2} = \sqrt{64 + 16 + 16} = \sqrt{96} = 4\sqrt{6} \]

### Step 4: Find the unit vector

The unit vector perpendicular to the plane is given by:

\[ \hat{n} = \frac{\vec{PQ} \times \vec{PR}}{|\vec{PQ} \times \vec{PR}|} \]

Substituting the values:

\[ \hat{n} = \frac{8\hat{i} + 4\hat{j} + 4\hat{k}}{4\sqrt{6}} = \frac{2\hat{i} + \hat{j} + \hat{k}}{\sqrt{6}} \]

### Final Answer

The unit vector perpendicular to the plane determined by the points \( P, Q, R \) is:

\[ \hat{n} = \frac{2\hat{i} + \hat{j} + \hat{k}}{\sqrt{6}} \]