<s>[INST] Intersection points of a Triangle and a Circle How can I find all intersection points of the following circle and triangle?
Triangle
$$A:=\begin{pmatrix}22\\-1.5\\1 \end{pmatrix} B:=\begin{pmatrix}27\\-2.25\\4 \end{pmatrix} C:=\begin{pmatrix}25.2\\-2\\4.7 \end{pmatrix}$$
Circle
$$\frac{9}{16}=(x-25)^2 + (y+2)^2 + (z-3)^2$$
What I did so far was to determine the line equations of the triangle (a, b and c):
$a : \overrightarrow {OX} = \begin{pmatrix}27\\-2.25\\4 \end{pmatrix}+ \lambda_1*\begin{pmatrix}-1.8\\0.25\\0.7 \end{pmatrix} $
$b : \overrightarrow {OX} = \begin{pmatrix}22\\-1.5\\1 \end{pmatrix}+ \lambda_2*\begin{pmatrix}3.2\\-0.5\\3.7 \end{pmatrix} $
$c : \overrightarrow {OX} = \begin{pmatrix}22\\-1.5\\1 \end{pmatrix}+ \lambda_3*\begin{pmatrix}5\\-0.75\\3 \end{pmatrix} $
But I am not sure what I have to do next...
 [/INST] If $\,\;\;\vec a+t\vec b\,,\;t\in\Bbb R\;\;$ is a line in $\,\Bbb R^3\,$, then each of its points can be expressed as
$$\begin{pmatrix}\alpha_1:=a_1+tb_1\\\alpha_2:=a_2+tb_2\\\alpha_3:=a_3+tb_3\end{pmatrix}\;,\;\;a_i,b_i,t\in\Bbb R$$
then a point as above belongs to the circle 
$$(x-25)^2+(y+2)^2+(z-3)^2=9\iff (\alpha_1-25)^2+(\alpha_2+2)^2+(\alpha_3-3)^2=9$$
so you have to substitute above, get a quadratic in $\,t\,$ and solve...of course, at most two different points for each straight line.
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