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

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

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$(4, 9, -3)$

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