Finding the Sphere Center and Plane Intersection

The question asks us to determine which plane contains the center of a sphere defined by the equation \(2x^2+2y^2 + 2z^2 + 3x + 3y+3z-6=0\). To solve this, we first need to find the coordinates of the sphere's center.

Deriving Sphere Center Coordinates

The standard equation of a sphere is given by:

\(x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0\)

In this form, the center of the sphere is located at the point \((-u, -v, -w)\).

Let's convert the given equation to this standard form. Divide the entire equation \(2x^2+2y^2 + 2z^2 + 3x + 3y+3z-6=0\) by 2:

\( \frac{2x^2}{2} + \frac{2y^2}{2} + \frac{2z^2}{2} + \frac{3x}{2} + \frac{3y}{2} + \frac{3z}{2} - \frac{6}{2} = 0 \)

This simplifies to:

\( x^2 + y^2 + z^2 + \frac{3}{2}x + \frac{3}{2}y + \frac{3}{2}z - 3 = 0 \)

Now, we compare this with the standard form \(x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0\):

• Comparing the coefficients of \(x\): \(2u = \frac{3}{2} \implies u = \frac{3}{4}\)
• Comparing the coefficients of \(y\): \(2v = \frac{3}{2} \implies v = \frac{3}{4}\)
• Comparing the coefficients of \(z\): \(2w = \frac{3}{2} \implies w = \frac{3}{4}\)
• The constant term is \(d = -3\).

Therefore, the center of the sphere is at:

\( \text{Center} = (-u, -v, -w) = \left(-\frac{3}{4}, -\frac{3}{4}, -\frac{3}{4}\right) \)

Checking Plane Intersection

A point lies on a plane if its coordinates satisfy the plane's equation. We will substitute the coordinates of the center, \(\left(-\frac{3}{4}, -\frac{3}{4}, -\frac{3}{4}\right)\), into each of the given plane equations.

Checking Option 1: \(2x+2y+2z – 3 = 0\)

Substitute the coordinates:

\( 2\left(-\frac{3}{4}\right) + 2\left(-\frac{3}{4}\right) + 2\left(-\frac{3}{4}\right) - 3 \)

\( = -\frac{6}{4} - \frac{6}{4} - \frac{6}{4} - 3 \)

\( = -\frac{18}{4} - 3 = -\frac{9}{2} - 3 = -\frac{15}{2} \)

Since \(-\frac{15}{2} \neq 0\), the center does not lie on this plane.

Checking Option 2: \(4x+4y+4z - 3 = 0\)

Substitute the coordinates:

\( 4\left(-\frac{3}{4}\right) + 4\left(-\frac{3}{4}\right) + 4\left(-\frac{3}{4}\right) - 3 \)

\( = -3 - 3 - 3 - 3 \)

\( = -12 \)

Since \(-12 \neq 0\), the center does not lie on this plane.

Checking Option 3: \(4x+8y+ 8z – 15 = 0\)

Substitute the coordinates:

\( 4\left(-\frac{3}{4}\right) + 8\left(-\frac{3}{4}\right) + 8\left(-\frac{3}{4}\right) - 15 \)

\( = -3 - 6 - 6 - 15 \)

\( = -30 \)

Since \(-30 \neq 0\), the center does not lie on this plane.

Checking Option 4: \(4x+8y+8z + 15 = 0\)

Substitute the coordinates:

\( 4\left(-\frac{3}{4}\right) + 8\left(-\frac{3}{4}\right) + 8\left(-\frac{3}{4}\right) + 15 \)

\( = -3 - 6 - 6 + 15 \)

\( = -15 + 15 \)

\( = 0 \)

Since the equation holds true (\(0 = 0\)), the center of the sphere lies on this plane.

Conclusion on Sphere Center Location

The center of the sphere lies on the plane \(4x+8y+8z + 15 = 0\).