Step 1: Understanding the Concept:
A plane touches a sphere (is tangent to the sphere) if and only if the perpendicular distance from the center of the sphere to the plane is equal to the radius of the sphere.
Step 2: Key Formula or Approach:
1. Find the center \((-u, -v, -w)\) and radius \(r = \sqrt{u^2+v^2+w^2-d}\) of the sphere from its equation \(x^2+y^2+z^2+2ux+2vy+2wz+d=0\).
2. The distance from a point \((x_0, y_0, z_0)\) to the plane \(Ax+By+Cz+D=0\) is \(d = \frac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}}\).
3. Set the distance from the center of the sphere to the plane equal to the radius and solve for \(k\).
Step 3: Detailed Explanation:
1. Find the center and radius of the sphere:
The equation of the sphere is \(x^2+y^2+z^2-2x+4y-6z+5=0\).
Comparing with the general form, we have:
\(2u = -2 \implies u = -1\)
\(2v = 4 \implies v = 2\)
\(2w = -6 \implies w = -3\)
\(d = 5\)
The center of the sphere is \((-u, -v, -w) = (1, -2, 3)\).
The radius of the sphere is:
\[ r = \sqrt{u^2+v^2+w^2-d} = \sqrt{(-1)^2 + 2^2 + (-3)^2 - 5} \]
\[ r = \sqrt{1+4+9-5} = \sqrt{9} = 3 \]
2. Find the distance from the center to the plane:
The equation of the plane is \(x-2y-2z=k \implies x-2y-2z-k=0\).
The center of the sphere is \((x_0, y_0, z_0) = (1, -2, 3)\).
The distance from the center to the plane is:
\[ \text{Distance} = \frac{|1(1) - 2(-2) - 2(3) - k|}{\sqrt{1^2 + (-2)^2 + (-2)^2}} = \frac{|1+4-6-k|}{\sqrt{1+4+4}} = \frac{|-1-k|}{\sqrt{9}} = \frac{|-(k+1)|}{3} = \frac{|k+1|}{3} \]
3. Set distance equal to radius:
For the plane to touch the sphere, this distance must be equal to the radius.
\[ \frac{|k+1|}{3} = 3 \]
\[ |k+1| = 9 \]
This gives two possible values for \(k\):
\(k+1 = 9 \implies k = 8\)
or
\(k+1 = -9 \implies k = -10\)
Checking the options, we see that -10 is listed.
Step 4: Final Answer:
The possible values for k are 8 and -10. Option (B) is -10.