Find the equation of the plane containing the lines `2x-y+z-3=0, 3x+y+z=5` and at a distance of `1/sqrt6` from the point `(2,1,-1)`.

Explanation

To find the equation of the plane containing the lines given by the equations \(2x - y + z - 3 = 0\) and \(3x + y + z - 5 = 0\), and at a distance of \(\frac{1}{\sqrt{6}}\) from the point \((2, 1, -1)\), we can follow these steps:

### Step 1: Write the equations of the lines
The equations of the lines are:
1. \(2x - y + z - 3 = 0\) (Equation 1)
2. \(3x + y + z - 5 = 0\) (Equation 2)

### Step 2: Formulate the equation of the plane
The general equation of a plane can be expressed as:
\[ Ax + By + Cz + D = 0 \]
To find the equation of the plane containing the two lines, we can express it as a linear combination of the two equations:
\[ (2x - y + z - 3) + \lambda(3x + y + z - 5) = 0 \]
Expanding this gives:
\[ (2 + 3\lambda)x + (-1 + \lambda)y + (1 + \lambda)z - (3 + 5\lambda) = 0 \]
This simplifies to:
\[ (2 + 3\lambda)x + (-1 + \lambda)y + (1 + \lambda)z - (3 + 5\lambda) = 0 \]

### Step 3: Identify coefficients
Let:
- \(A = 2 + 3\lambda\)
- \(B = -1 + \lambda\)
- \(C = 1 + \lambda\)
- \(D = -(3 + 5\lambda)\)

### Step 4: Use the distance formula
The distance \(d\) from a point \((x_0, y_0, z_0)\) to a plane \(Ax + By + Cz + D = 0\) is given by:
\[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
Substituting the point \((2, 1, -1)\) into the distance formula, we set \(d = \frac{1}{\sqrt{6}}\):
\[ \frac{|A(2) + B(1) + C(-1) + D|}{\sqrt{A^2 + B^2 + C^2}} = \frac{1}{\sqrt{6}} \]
This leads to:
\[ |2A + B - C + D| = \frac{1}{\sqrt{6}} \sqrt{A^2 + B^2 + C^2} \]

### Step 5: Substitute \(A\), \(B\), \(C\), and \(D\)
Substituting the values of \(A\), \(B\), \(C\), and \(D\):
\[ |2(2 + 3\lambda) + (-1 + \lambda) - (1 + \lambda) - (3 + 5\lambda)| = \frac{1}{\sqrt{6}} \sqrt{(2 + 3\lambda)^2 + (-1 + \lambda)^2 + (1 + \lambda)^2} \]

### Step 6: Simplify and solve for \(\lambda\)
Simplifying the left-hand side:
\[ |4 + 6\lambda - 1 + \lambda - 1 - \lambda - 3 - 5\lambda| = |(4 - 5) + (6 + 1 - 1 - 5)\lambda| = |(-1) + (1)\lambda| = |-\lambda - 1| \]
The right-hand side becomes:
\[ \frac{1}{\sqrt{6}} \sqrt{(2 + 3\lambda)^2 + (-1 + \lambda)^2 + (1 + \lambda)^2} \]
This gives us a quadratic equation in \(\lambda\) which we can solve.

### Step 7: Solve for \(\lambda\)
After simplifying, we will find two values for \(\lambda\):
1. \(\lambda = 0\)
2. \(\lambda = -\frac{24}{5}\)

### Step 8: Find the equations of the planes
Substituting these values back into the equation of the plane:
1. For \(\lambda = 0\):
\[ 2x - y + z - 3 = 0 \]
2. For \(\lambda = -\frac{24}{5}\):
\[ 62x + 29y + 19z - 105 = 0 \]

### Final Answer
The equations of the planes are:
1. \(2x - y + z - 3 = 0\)
2. \(62x + 29y + 19z - 105 = 0\)