Step 1: Conceptual Basis:
A plane is tangent to a sphere if and only if the perpendicular distance from the sphere's center to the plane equals the sphere's radius.
Step 2: Methodology:
1. Determine the sphere's center (C) and radius (R) from its equation.2. Employ the formula for the distance (d) from a point \((x_0, y_0, z_0)\) to a plane \(Ax + By + Cz + D = 0\), which is \(d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}\).3. Equate \(d\) to \(R\) (\(d = R\)) and solve for the unknown parameter \(\lambda\).
Step 3: Detailed Calculation:
1. Sphere Characteristics:
Given sphere equation: \(x^2 + y^2 + z^2 - 2x - 2y - 2z - 6 = 0\).Completing the square yields:\[ (x^2 - 2x + 1) + (y^2 - 2y + 1) + (z^2 - 2z + 1) - 6 - 1 - 1 - 1 = 0 \]\[ (x-1)^2 + (y-1)^2 + (z-1)^2 = 9 \]Sphere center: \(C = (1, 1, 1)\).Sphere radius: \(R = \sqrt{9} = 3\).
2. Distance to Plane:
Plane equation: \(x + y + z - \sqrt{3}\lambda = 0\).Point (center): \((x_0, y_0, z_0) = (1, 1, 1)\).The distance \(d\) is calculated as:\[ d = \frac{|1(1) + 1(1) + 1(1) - \sqrt{3}\lambda|}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{|3 - \sqrt{3}\lambda|}{\sqrt{3}} \]3. Solving for \(\lambda\):
Set distance equal to radius: \(d = R\).\[ \frac{|3 - \sqrt{3}\lambda|}{\sqrt{3}} = 3 \]\[ |3 - \sqrt{3}\lambda| = 3\sqrt{3} \]This leads to two possibilities:Case 1: \( 3 - \sqrt{3}\lambda = 3\sqrt{3} \)\[ \sqrt{3}\lambda = 3 - 3\sqrt{3} \]\[ \lambda = \frac{3}{\sqrt{3}} - 3 = \sqrt{3} - 3 \]Case 2: \( 3 - \sqrt{3}\lambda = -3\sqrt{3} \)\[ \sqrt{3}\lambda = 3 + 3\sqrt{3} \]\[ \lambda = \frac{3}{\sqrt{3}} + 3 = \sqrt{3} + 3 \]Thus, the possible values for \(\lambda\) are \( \sqrt{3} - 3 \) and \( \sqrt{3} + 3 \), which can be expressed as \( \lambda = \sqrt{3} \pm 3 \).
Step 4: Conclusion:
The determined values for \(\lambda\) are \( \sqrt{3} \pm 3 \).