Find the value of `lambda` for which the plane `x+y+z=sqrt(3)lambda` touches the sphere `x^2+y^2+z^2-2x-2y-2z=6.`

Answer

To find the value of `lambda` for which the plane \( x + y + z = \sqrt{3} \lambda \) touches the sphere defined by the equation \( x^2 + y^2 + z^2 - 2x - 2y - 2z = 6 \), we can follow these steps:

### Step 1: Rewrite the Sphere Equation
The given equation of the sphere can be rearranged into standard form. We start with:
\[ x^2 + y^2 + z^2 - 2x - 2y - 2z = 6 \]
We can complete the square for each variable:
\[ (x^2 - 2x) + (y^2 - 2y) + (z^2 - 2z) = 6 \]
This becomes:
\[ (x - 1)^2 - 1 + (y - 1)^2 - 1 + (z - 1)^2 - 1 = 6 \]
Simplifying gives:
\[ (x - 1)^2 + (y - 1)^2 + (z - 1)^2 = 9 \]
Thus, the center of the sphere is \( (1, 1, 1) \) and the radius is \( 3 \).

### Step 2: Rewrite the Plane Equation
The plane can be rewritten as:
\[ x + y + z - \sqrt{3} \lambda = 0 \]

### Step 3: Find the Distance from the Center of the Sphere to the Plane
The distance \( d \) from a point \( (x_0, y_0, z_0) \) to the plane \( Ax + By + Cz + D = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
For our plane, \( A = 1, B = 1, C = 1, D = -\sqrt{3} \lambda \), and the center of the sphere \( (1, 1, 1) \):
\[ d = \frac{|1 \cdot 1 + 1 \cdot 1 + 1 \cdot 1 - \sqrt{3} \lambda|}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{|3 - \sqrt{3} \lambda|}{\sqrt{3}} \]

### Step 4: Set the Distance Equal to the Radius
For the plane to touch the sphere, the distance must equal the radius:
\[ \frac{|3 - \sqrt{3} \lambda|}{\sqrt{3}} = 3 \]
Multiplying both sides by \( \sqrt{3} \):
\[ |3 - \sqrt{3} \lambda| = 3\sqrt{3} \]

### Step 5: Solve the Absolute Value Equation
This gives us two cases to consider:
1. \( 3 - \sqrt{3} \lambda = 3\sqrt{3} \)
2. \( 3 - \sqrt{3} \lambda = -3\sqrt{3} \)

#### Case 1:
\[ 3 - \sqrt{3} \lambda = 3\sqrt{3} \]
Solving for \( \lambda \):
\[ -\sqrt{3} \lambda = 3\sqrt{3} - 3 \]
\[ \lambda = -\frac{3\sqrt{3} - 3}{\sqrt{3}} = 3 + \sqrt{3} \]

#### Case 2:
\[ 3 - \sqrt{3} \lambda = -3\sqrt{3} \]
Solving for \( \lambda \):
\[ -\sqrt{3} \lambda = -3\sqrt{3} - 3 \]
\[ \lambda = -\frac{-3\sqrt{3} - 3}{\sqrt{3}} = \sqrt{3} - 3 \]

### Final Result
The values of \( \lambda \) for which the plane touches the sphere are:
\[ \lambda = 3 + \sqrt{3} \quad \text{and} \quad \lambda = \sqrt{3} - 3 \]