Find an equation for the sphere centered at $(3,3,-1)$ and passing through $(1,3,3)$. 1. $3 x^{2}+3 y^{2}-z^{2}+1=0$ 2. $3 x^{2}+3 y^{2}-z^{2}=1$ 3. $x^{2}+y^{2}+z^{2}+6 x+6 y-2 z=1$ 4. $x^{2}+y^{2}+z^{2}-6 x-6 y+2 z=1$ 5. $x^{2}+y^{2}+z^{2}-6 x-6 y+2 z+1=0$ 6. $x^{2}+y^{2}+z^{2}+6 x+6 y-2 z+1=0$

Step 1 ：The equation of a sphere in 3D space is given by \((x-a)^2 + (y-b)^2 + (z-c)^2 = r^2\), where \((a,b,c)\) is the center of the sphere and \(r\) is the radius.

Step 2 ：We know the center of the sphere is \((3,3,-1)\), so we can substitute these values into the equation to get \((x-3)^2 + (y-3)^2 + (z+1)^2 = r^2\).

Step 3 ：We also know that the sphere passes through the point \((1,3,3)\), so we can substitute these values into the equation to solve for \(r^2\).

Step 4 ：Substituting \((1,3,3)\) into the equation, we get \((1-3)^2 + (3-3)^2 + (3+1)^2 = r^2\), which simplifies to \(4 + 0 + 16 = r^2\), so \(r^2 = 20\).

Step 5 ：Finally, we can substitute the value of \(r^2\) back into the equation to get the final equation of the sphere: \((x-3)^2 + (y-3)^2 + (z+1)^2 = 20\).

Step 6 ：Simplifying this equation, we get \(x^{2}+y^{2}+z^{2}-6 x-6 y+2 z+1=0\).

Step 7 ：Final Answer: The equation of the sphere is \(\boxed{x^{2}+y^{2}+z^{2}-6 x-6 y+2 z+1=0}\).