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
 A: The sphere has for equation $$S \equiv (x+3)^2-9+y^2+(z-1)^2-1+1=(x+3)^2+y^2+(z-1)^2-9=0$$ which means that its center is $C= (-3,0,1)$ and its radius is equal to $r=3$. The line has for equations $$L \equiv \begin{cases}
x+z=16\\
2y-3z=-30
\end{cases}$$ It goes through the point $P_L=(16,-15,0)$ and has $v_L=(-2,3,2)$ as a direction vector.
The general equation of a plane $\mathscr P$ is $a x+by+cz+d=0$ where $a^2+b^2+c^2=1$. $\mathscr P$ is passing through $P_L$ so you must have $16a-15b+d=0$. And $\mathscr P$ contains the direction of the line, so $-2a+3b+2c=0$. Finally as $\mathscr P$ is supposed to be tangent to the sphere, the distance of $\mathscr P$ to $C$ is equal to the radius. Hence $\vert -3a+c+d \vert = 3$.
Therefore, you have to solve following system of equations $$\begin{cases}
a^2+b^2+c^2=1\\
16a-15b+d=0\\
-2a+3b+2c=0\\
\vert -3a+c+d \vert = 3
\end{cases}$$
You'll get two different solutions depending on the sign of $\vert -3a+c+d \vert$.
A: 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 $\color{red}{-x-2y+2z=14}$.
A: Hint.
Given a plane 
$$
\Pi\to (p-p_0)\cdot \vec n = 0
$$
and a line
$$
l\to p = p_1 + \lambda \vec v
$$
with $p_0\ne p_1$. If $l\in \Pi$ then the conditions
$$
(p_1+\lambda\vec v-p_0)\cdot \vec n = 0\Rightarrow \cases{(p_1-p_0)\cdot \vec n=0\\
\vec v\cdot \vec n = 0\\
||\vec n||^2= 1}\ \ \ \ \ \ \ \ (1)
$$
define the $\vec n$ vector regarding $l\in \Pi$.
Now if we have a sphere 
$$
S\to ||p-p_2||^2=r^2
$$
such that $p_0\in S$ is a tangent point between $S$ and $\Pi$ then $p_0-p_2 = r\vec n$ and then
$$
p_0 = p_2+r\vec n\ \ \ \ \ \ \ \ (2)
$$
Concluding, any line in $\Pi$ non parallel to $l$ and passing by $p_0$ is a solution.
NOTE
Here
$$
p_1 = (6,-15,0),\ \ \vec v = \left(-\frac 13,\frac 12,\frac 13\right),\ \ \ p_2 = (-3,0,1),\ \ \ r=3
$$
and from $(1)$ and $(2)$
$$
\cases{(p_1-p_2)\cdot \vec n=r\\
\vec v\cdot \vec n = 0\\
||\vec n||^2= 1}
$$
define finally $\vec n$
