If (2, 3, 5) is one end of a diameter of the | vedclass

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(B) The equation of the sphere is ${x^2} + {y^2} + {z^2} - 6x - 12y - 2z + 20 = 0$. Comparing this with the general form ${x^2} + {y^2} + {z^2} + 2ux

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$(4, 9, -3)$

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(B) The equation of the sphere is ${x^2} + {y^2} + {z^2} - 6x - 12y - 2z + 20 = 0$. Comparing this with the general form ${x^2} + {y^2} + {z^2} + 2ux + 2vy + 2wz + d = 0$,we get the center of the sphere as $(-u, -v, -w) = (3, 6, 1)$. Let the coordinates of the other end of the diameter be $(x, y, z)$. Since the center of the sphere is the midpoint of the diameter,we have: $\frac{2 + x}{2} = 3 \implies 2 + x = 6 \implies x = 4$ $\frac{3 + y}{2} = 6 \implies 3 + y = 12 \implies y = 9$ $\frac{5 + z}{2} = 1 \implies 5 + z = 2 \implies z = -3$ Thus,the coordinates of the other end are $(4, 9, -3)$.

If $(2, 3, 5)$ is one end of a diameter of the sphere ${x^2} + {y^2} + {z^2} - 6x - 12y - 2z + 20 = 0$,then the coordinates of the other end of the diameter are:

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