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 will follow these steps:

### Step 1: Rewrite the equation of the sphere

The equation of the sphere can be rewritten in standard form. We start with:

\[ x^2 + y^2 + z^2 - 2x - 2y - 2z = 6 \]

We can complete the square for each variable:

\[ (x^2 - 2x) + (y^2 - 2y) + (z^2 - 2z) = 6 \]

Completing the square:

\[ (x - 1)^2 - 1 + (y - 1)^2 - 1 + (z - 1)^2 - 1 = 6 \]

This simplifies to:

\[ (x - 1)^2 + (y - 1)^2 + (z - 1)^2 = 9 \]

Thus, the center of the sphere is \( (1, 1, 1) \) and the radius is \( 3 \).

### Step 2: Find the distance from the center of the sphere to the plane

The equation of the plane is given by:

\[ x + y + z = \sqrt{3} \lambda \]

We can express this in the form \( Ax + By + Cz + D = 0 \):

\[ x + y + z - \sqrt{3} \lambda = 0 \]

Here, \( A = 1, B = 1, C = 1, D = -\sqrt{3} \lambda \).

The distance \( d \) from a point \( (x_0, y_0, z_0) \) to the plane \( Ax + By + Cz + D = 0 \) is given by:

\[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]

Substituting 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 3: Set the distance equal to the radius of the sphere

Since the plane touches the sphere, the distance \( d \) must equal the radius of the sphere, which is \( 3 \):

\[ \frac{|3 - \sqrt{3} \lambda|}{\sqrt{3}} = 3 \]

Multiplying both sides by \( \sqrt{3} \):

\[ |3 - \sqrt{3} \lambda| = 3\sqrt{3} \]

### Step 4: 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:**

\[ 3 - \sqrt{3} \lambda = 3\sqrt{3} \]
\[ -\sqrt{3} \lambda = 3\sqrt{3} - 3 \]
\[ -\sqrt{3} \lambda = 3(\sqrt{3} - 1) \]
\[ \lambda = -\frac{3(\sqrt{3} - 1)}{\sqrt{3}} = -3 + \sqrt{3} \]

**Case 2:**

\[ 3 - \sqrt{3} \lambda = -3\sqrt{3} \]
\[ -\sqrt{3} \lambda = -3\sqrt{3} - 3 \]
\[ -\sqrt{3} \lambda = -3(\sqrt{3} + 1) \]
\[ \lambda = \frac{3(\sqrt{3} + 1)}{\sqrt{3}} = 3 + \sqrt{3} \]

### Final Answer

Thus, the values of \( \lambda \) for which the plane touches the sphere are:

\[ \lambda = -3 + \sqrt{3} \quad \text{and} \quad \lambda = 3 + \sqrt{3} \]