Hello, SyNtHeSiS!

$\displaystyle \text{Find the point on the sphere }\,(x - 2)^2 + (y - 2)^2 + (z-7)^2 \:= \:99$

$\displaystyle \text{which is furthest from the point }\,C(1,3,4).$

Code:

* * * o C
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* ♥ P
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* o *
* * A *
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Q ♥ *
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Construct a line through $\displaystyle \,C$ and the center of the sphere $\displaystyle \,A.$

It will intersect the sphere in two points, $\displaystyle \,P$ and $\displaystyle Q.$

. . We want $\displaystyle Q$, the point further from $\displaystyle \,C.$

The line from $\displaystyle C(1,3,4)$ to $\displaystyle A(2,2,7)$ has direction $\displaystyle \langle 1,-1,3\rangle$

The line has parametric equations: .$\displaystyle \begin{Bmatrix}x &=& 1 + t \\ y &=& 3 - t \\ z &=& 4+3t \end{Bmatrix}$

Substitute into the equation of the sphere:

. . $\displaystyle \left[(1-t) - 2\right]^2 + \left[(3+t)-2\right]^2 + \left[([4-3t)-7\right]^2 \:=\:99$

. . $\displaystyle 11t^2 + 22t + 11 \:=\:99 \quad\Rightarrow\quad t^2 + 2t + 1 \:=\:9$

. . $\displaystyle t^2 + 2t - 8 \:=\:0 \quad\Rightarrow\quad (t-2)(t+4)\:=\:0 \quad\Rightarrow\quad t \:=\:2,\:-4 $

$\displaystyle \begin{array}{ccccccc}t = 2\!: & (x,y,z) &=& (\text{-}1,5,\text{-}2) & \text{point J} \\

t = \text{-}4\!: & (x,y,z) &=& (5,\text{-}1,16) & \text{point K}\end{array}$

I suspect that point $\displaystyle \,K$ is further from $\displaystyle \,C$, but let's make sure.

Distance from $\displaystyle C(1,3,4)$ to $\displaystyle J(\text{-}1,5,\text{-}2)\!:$

. . $\displaystyle \overline{CJ} \;=\;\sqrt{(\text{-}1-1)^2 + (5-3)^3 + (\text{-}2-4)^2} \;=\;\sqrt{44} $

Distance from $\displaystyle C(1,3,4)$ to $\displaystyle K(5,\text{-}1,16)\!:$

. . $\displaystyle CK \;=\;\sqrt{(5-1)^2 + (\text{-}1-3)^2 + (16-4)^2} \;=\;\sqrt{176}$

Hence, $\displaystyle K$ is the point we are seeking.

Therefore: .$\displaystyle Q(5,\text{-}1,16)$