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If $(2,7,3)$ is one end of a diameter of the sphere $x^{2}+y^{2}+z^{2}-6 x-12 y-2 z+20=0$, then the coordinates of the other end of the diameter are
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The correct answer is:
$(4,5,-1)$
Given sphere is
$\begin{array}{l}
x^{2}+y^{2}+z^{2}-6 x-12 y-2 z+20=0 \\
\text { Centre } \equiv(3,6,1)
\end{array}$
Here, one end of diameter is $(2,7,3)$.
Let the other end of the diameter be $(x, y, z)$
Centre of the sphere will be the mid-point of the ends of diameter.
So, $(3,6,1)=\left(\frac{2+x}{2}, \frac{7+y}{2}, \frac{3+z}{2}\right)$
$\Rightarrow \quad 2+\mathrm{x}=6 \Rightarrow \mathrm{x}=4$ $\Rightarrow \quad 7+\mathrm{y}=12 \Rightarrow \mathrm{y}=5$ and $3+\mathrm{z}=2 \Rightarrow \mathrm{z}=-$
and $3+z=2 \Rightarrow z=-1,5,-1)$
$\begin{array}{l}
x^{2}+y^{2}+z^{2}-6 x-12 y-2 z+20=0 \\
\text { Centre } \equiv(3,6,1)
\end{array}$
Here, one end of diameter is $(2,7,3)$.
Let the other end of the diameter be $(x, y, z)$
Centre of the sphere will be the mid-point of the ends of diameter.
So, $(3,6,1)=\left(\frac{2+x}{2}, \frac{7+y}{2}, \frac{3+z}{2}\right)$
$\Rightarrow \quad 2+\mathrm{x}=6 \Rightarrow \mathrm{x}=4$ $\Rightarrow \quad 7+\mathrm{y}=12 \Rightarrow \mathrm{y}=5$ and $3+\mathrm{z}=2 \Rightarrow \mathrm{z}=-$
and $3+z=2 \Rightarrow z=-1,5,-1)$
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