<s>[INST] Find the equations of the tangent planes to the sphere $x^2+y^2+z^2+2x-4y+6z-7=0,$ which intersect in the line $6x-3y-23=0=3z+2.$ Find the equations of the tangent planes to the sphere $x^2+y^2+z^2+2x-4y+6z-7=0,$ which intersect in the line $6x-3y-23=0=3z+2.$

Let the tangent planes be $A_1x+B_1y+C_1z+D_1=0$ and $A_2x+B_2y+C_2z+D_2=0$
As the line $6x-3y-23=0=3z+2$ lies on both the planes,so put $z=\frac{-2}{3},x=\frac{3y+23}{6}$ in the both equations of the planes,
$A_1\frac{3y+23}{6}+B_1y-\frac{2C_1}{3}+D_1=0$ and $A_2\frac{3y+23}{6}+B_2y-\frac{2C_2}{3}+D_2=0$
I am stuck here.Please help.
The answer given in the book is $2x-y+4z-5=0$ and $4x-2y-z=16.$
 [/INST] Given sphere:$$
x^2+y^2+z^2+2x-4y+6z-7=0.
$$
Required planes which intersect in the given line are given by$$
6x-3y-23+λ(3z+2)= 0. \tag{1}
$$
The plane (1) will be tangent to the given sphere when radius of the sphere is equal to the perpendicular distance of the plane from the center $(-1, 2, -3)$ of the sphere. This gives $λ = -\frac{1}{2}, 4$.
With these values of $λ$, equation (1) gives required tangent planes.
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