<s>[INST] find the tangent to the sphere obtain the equations of tangent to sphere
$$x^2+ y^2+z^2+6x-2z+1 = 0$$
which pass through the line
$$3 (16-x) = 3z=2y+30$$
Now I know if the plane is $$lx +my+n z=p$$
then $$-I/3 +m/2+n/3=0$$
also $(16,-15,0)$ is a point on the plane
I know that there is $2$ answers , but how to proceed
 [/INST] Let $R(a,b,c)$ be the point of tangency, and let $P(16, -15, 0)$ and $Q(6, 0,10)$ be two points on the line.
Since the center of the sphere is $C(-3, 0, 1)$, a normal vector to the plane is given by $\vec{n}=\langle a+3,b,c-1\rangle$.
Therefore the plane has equation $(a+3)(x-a)+b(y-b)+(c-1)(z-c)=0$, which gives 
$\;\;\color{blue}{(a+3)x+by+(c-1)z=-3a+c-1}\;\;\;$ since $a^2+b^2+c^2=-6a+2c-1$.
Substituting the coordinates of $Q$ gives $6(a+3)+10(c-1)=-3a+c-1$, 
so $a+c=-1$ and $\color{blue}{a=-c-1}$.
Substituting the coordinates of $P$ gives $16(a+3)-15b=-3a+c-1$, 
so $3b+4c=6$ and $\color{blue}{b=2-\frac{4}{3}c}$.
Substituting into $a^2+b^2+c^2+6a-2c+1=0$ gives 
$(c+1)^2+(2-\frac{4}{3}c)^2+c^2+6(-c-1)-2c+1=0$,
so $\frac{34}{9}c^2-\frac{34}{3}c=0$ and thus $c^2-3c=c(c-3)=0$.
If $c=0$, $a=-1$ and $b=2$; so the plane has equation $\color{red}{2x+2y-z=2}$.
If $c=3$, $a=-4$ and $b=-2$; so the plane has equation $\color{red}{-x-2y+2z=14}$.
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