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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4x²+4y²+4z²-12x+8y-16z+13=0,

4x²-12x+9+4y²+8y+4+4z²-16z+16-29+13=0,

4(x-3/2)²+4(y+1)²+4(z-2)²-16=0,

(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:

x+y+z=5/2.

ago by Top Rated User (823k points)

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