The example I have for this problem is the following: What are the tangent planes of the sphere $x^2 + y^2 + z^2 = 36$, which contains the line $\left(\begin{array}{c}x\\ y\\ z\end{array}\right) = \left(\begin{array}{c}6\\ 0\\ 3\end{array}\right) + \lambda\left(\begin{array}{c}1\\ 2\\ 0\end{array}\right)$?
I haven't been able to come up with any way to do this.
You can reduce this to a two-dimensional problem by considering the plane which passes through the sphere’s center and is orthogonal to the line. The tangent lines that you get, together with the original line, define the required tangent planes.
For this problem, that’s pretty straightforward since the quadric surface (the sphere) is nicely symmetric, so there aren’t really any rotations to deal with, and it’s already centered at the origin. For the general problem, though, I think an easier approach is to find the pencil of planes that pass through the line, then apply the constraint that the planes must be tangent to the quadric. If the line is given in the form $\mathbf p+t\mathbf v$, you can do this by working in homogeneous coordinates: First, find a basis for the null space of the matrix $$\begin{bmatrix}\mathbf p^T & 1 \\ \mathbf v^T & 0 \end{bmatrix}.$$ Any two points on the line will do for the rows of this matrix, but since the direction vector in the parametric equation of the line corresponds to a point at infinity that lies on the line, we can read the above two points directly from that equation. This basis will consist of two planes, $\mathbf\pi_1$ and $\mathbf\pi_2$, that contain the line, and so every plane that does is of the form $\mathbf\pi = \lambda\pi_1+\mu\pi_2$. For this plane to be tangent to the quadric represented by the matrix $C$, it must satisfy the dual equation $\mathbf\pi^TC^{-1}\mathbf\pi=(\lambda\pi_1+\mu\pi_2)^TC^{-1}(\lambda\pi_1+\mu\pi_2)=0$. Solve the resulting quadratic equation for $\lambda$ and $\mu$.
Applying this method to your example, we have $C = \operatorname{diag}(1,1,1,-36)$, so $C^{-1}=\operatorname{diag}(1,1,1,-1/36)$. A basis for the null space of $$\begin{bmatrix}6&0&3&1\\1&2&0&0\end{bmatrix}$$ consists of $(-2,1,0,12)^T$ and $(-2,1,4,0)^T$ (found via row-reduction), so the general plane through the line is $\mathbf\pi = [-2(\lambda+\mu):\lambda+\mu:4\mu:12\lambda]$. The resulting equation for the tangent condition is $$\lambda^2+10\lambda\mu+21\mu^2=0,$$ which factors nicely into $$(\lambda+3\mu)(\lambda+7\mu)=0.$$ Therefore, setting $\mu=-1$ and $\lambda=3$ in $\mathbf\pi$ will give one of the tangent planes, and $\mu=-1$, $\lambda=7$ the other. These substitutions produce the homogeneous vectors $[-4:2:-4:36]$ and $[-12:6:-4:84]$, which correspond to the inhomogeneous Cartesian equations $$2x-y+2z=18 \\ 6x-3y+2z=42.$$ (I’ve divided through by $2$ to reduce the magnitudes of the coefficients.)