Step 1: Understanding the Concept:
We need to find the equation of a sphere that satisfies four conditions: passing through three given points and having its center on a given plane.
Step 2: Key Formula or Approach:
1. Let the general equation of the sphere be \(x^2+y^2+z^2+2ux+2vy+2wz+d=0\).
2. Substitute the coordinates of the three given points into this equation to get three linear equations in u, v, w, and d.
3. The center of the sphere is \((-u, -v, -w)\). This point lies on the plane \(x+y+z=6\), which gives a fourth linear equation: \(-u-v-w=6\).
4. Solve the system of four linear equations to find u, v, w, and d.
5. Substitute these values back into the general equation of the sphere.
Step 3: Detailed Explanation:
Let the equation of the sphere be \(x^2+y^2+z^2+2ux+2vy+2wz+d=0\).
The sphere passes through the following points:
Point (1,0,0):
\[ (1)^2+0+0+2u(1)+0+0+d=0 \implies 1+2u+d=0 \quad (1) \]
Point (0,1,0):
\[ 0+(1)^2+0+0+2v(1)+0+d=0 \implies 1+2v+d=0 \quad (2) \]
Point (0,0,1):
\[ 0+0+(1)^2+0+0+2w(1)+d=0 \implies 1+2w+d=0 \quad (3) \]
From (1), (2), and (3), we can see that:
\[ 1+2u+d = 1+2v+d \implies 2u=2v \implies u=v \]
\[ 1+2v+d = 1+2w+d \implies 2v=2w \implies v=w \]
So, we have \(u=v=w\).
The center of the sphere is \((-u,-v,-w)\). This point lies on the plane \(x+y+z=6\).
\[ (-u) + (-v) + (-w) = 6 \implies -u-v-w = 6 \]
Since \(u=v=w\), we can substitute:
\[ -u-u-u = 6 \implies -3u = 6 \implies u = -2 \]
Therefore, \(u=v=w=-2\).
Now, use equation (1) to find d:
\[ 1+2u+d=0 \implies 1+2(-2)+d=0 \implies 1-4+d=0 \implies -3+d=0 \implies d=3 \]
We have found all the coefficients: \(u=-2, v=-2, w=-2, d=3\).
Substitute these into the general equation of the sphere:
\[ x^2+y^2+z^2+2(-2)x+2(-2)y+2(-2)z+3=0 \]
\[ x^2+y^2+z^2-4x-4y-4z+3=0 \]
Step 4: Final Answer:
The equation of the sphere is \(x^2+y^2+z^2-4x-4y-4z+3=0\), which corresponds to option (D).