Determine the equation of the plane that contains center of the sphere 4x²+4y²+4x²-12x+8y-16z+13=0 and is parallel to the plane that contains the points (0,1,1), (1,0,1), and (1,1,0).
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(x-3/2)²+(y+1)²+(z-2)²=4² is a sphere of radius 4, centre (3/2,-1,2).

The plane containing the given points can be represented:

Ax+By+Cz+D=0. Plugging in the points:

B+C+D=0, A+C+D=0, A+B+D=0, where A, B, C, D are constants.

We can write this system as:

B+C=A+C=A+B=-D. Therefore A=B=C=-D/2 and the equation of the plane is:

(-x-y-z)/2+1=0, x+y+z=2.

All parallel planes will have the form: x+y+z=p where p is a constant.

To pass through the centre of the sphere:

3/2-1+2=5/2=p, making the equation of the required plane:


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