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

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 $\vert -3a+c+d \vert$.
