If (2,3,-3) is one end of a diameter of the s | 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 equation of a sphere $x^2+y^2+z^2+2ux+2vy+2wz+d=0$,we

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$(4,9,5)$

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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 equation of a sphere $x^2+y^2+z^2+2ux+2vy+2wz+d=0$,we get $2u=-6, 2v=-12, 2w=-2$. Thus,$u=-3, v=-6, w=-1$. The centre of the sphere is $(-u,-v,-w) = (3,6,1)$. Let the given end of the diameter be $A = (2,3,-3)$ and the other end be $B = (\alpha, \beta, \gamma)$. Since the centre $O(3,6,1)$ is the midpoint of the diameter $AB$,we have: $O = \left( \frac{\alpha+2}{2}, \frac{\beta+3}{2}, \frac{\gamma-3}{2} \right) = (3,6,1)$. Equating the coordinates: $\frac{\alpha+2}{2} = 3 \Rightarrow \alpha+2 = 6 \Rightarrow \alpha = 4$. $\frac{\beta+3}{2} = 6 \Rightarrow \beta+3 = 12 \Rightarrow \beta = 9$. $\frac{\gamma-3}{2} = 1 \Rightarrow \gamma-3 = 2 \Rightarrow \gamma = 5$. Therefore,the other end of the diameter is $(4,9,5)$.

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

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