Let `A(1,1,1),B(2,3,5)a n dC(-1,0,2)` be three points, then equation of a plane parallel to the plane `A B C` which is at distance 2units is

Explanation

To find the equation of a plane parallel to the plane formed by points A, B, and C, which is at a distance of 2 units, we can follow these steps:

### Step 1: Find the equation of the plane ABC
We start by determining the vectors AB and AC:

- \( A(1, 1, 1) \)
- \( B(2, 3, 5) \)
- \( C(-1, 0, 2) \)

The vector \( \overrightarrow{AB} \) is given by:
\[ \overrightarrow{AB} = B - A = (2 - 1, 3 - 1, 5 - 1) = (1, 2, 4) \]

The vector \( \overrightarrow{AC} \) is given by:
\[ \overrightarrow{AC} = C - A = (-1 - 1, 0 - 1, 2 - 1) = (-2, -1, 1) \]

Next, we find the normal vector \( \overrightarrow{n} \) to the plane ABC by taking the cross product \( \overrightarrow{AB} \times \overrightarrow{AC} \).

Calculating the cross product:
\[ \overrightarrow{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 4 \\ -2 & -1 & 1 \end{vmatrix} \]

Calculating the determinant:
\[ \overrightarrow{n} = \hat{i}(2 \cdot 1 - 4 \cdot (-1)) - \hat{j}(1 \cdot 1 - 4 \cdot (-2)) + \hat{k}(1 \cdot (-1) - 2 \cdot (-2)) \]
\[ = \hat{i}(2 + 4) - \hat{j}(1 + 8) + \hat{k}(-1 + 4) \]
\[ = 6\hat{i} - 9\hat{j} + 3\hat{k} \]
Thus, the normal vector \( \overrightarrow{n} = (6, -9, 3) \).

### Step 2: Equation of the plane ABC
Using the point-normal form of the plane equation:
\[ 6(x - 1) - 9(y - 1) + 3(z - 1) = 0 \]
Expanding this:
\[ 6x - 6 - 9y + 9 + 3z - 3 = 0 \]
\[ 6x - 9y + 3z = 0 \]
This simplifies to:
\[ 2x - 3y + z = 1 \]

### Step 3: Find the equation of the parallel plane
The general form of the equation of a plane parallel to the original plane is:
\[ 2x - 3y + z + k = 0 \]

### Step 4: Calculate the distance between the two planes
The distance \( d \) between two parallel planes \( Ax + By + Cz + D_1 = 0 \) and \( Ax + By + Cz + D_2 = 0 \) is given by:
\[ d = \frac{|D_2 - D_1|}{\sqrt{A^2 + B^2 + C^2}} \]
Here, \( D_1 = 1 \) and \( D_2 = k \). Thus:
\[ d = \frac{|k - 1|}{\sqrt{2^2 + (-3)^2 + 1^2}} = \frac{|k - 1|}{\sqrt{4 + 9 + 1}} = \frac{|k - 1|}{\sqrt{14}} \]
Setting this equal to 2:
\[ \frac{|k - 1|}{\sqrt{14}} = 2 \]
Multiplying both sides by \( \sqrt{14} \):
\[ |k - 1| = 2\sqrt{14} \]

### Step 5: Solve for \( k \)
This gives us two cases:
1. \( k - 1 = 2\sqrt{14} \) which gives \( k = 2\sqrt{14} + 1 \)
2. \( k - 1 = -2\sqrt{14} \) which gives \( k = -2\sqrt{14} + 1 \)

### Step 6: Write the equations of the two parallel planes
Thus, the equations of the planes parallel to ABC at a distance of 2 units are:
1. \( 2x - 3y + z + (2\sqrt{14} + 1) = 0 \)
2. \( 2x - 3y + z + (-2\sqrt{14} + 1) = 0 \)

### Final Answer
The equations of the planes are:
1. \( 2x - 3y + z + (2\sqrt{14} + 1) = 0 \)
2. \( 2x - 3y + z + (-2\sqrt{14} + 1) = 0 \)