Equation of plane containing the line and lying at a distance from the point I need to find the equation(s) of the plane containing the line
$$
\left\{ 
\begin{array}{c}
x=1+k \\
y=1 \\
z = -2k
\end{array}
\right. 
$$
lying at distance 1 from the point (-1,0,1).
I was trying to use "Distance from point to a plane" formula and compose a system, given that I know collinear points on the plane (from the line), however, I apparently miss something to finalize the solution.
 A: There is a one-parameter family of planes that all include the given line. You can use the distance formula with this to generate a one-variable quadratic equation.  
There are several ways to produce this family of planes. For instance, note that their normals must be perpendicular to the line. We can read a direction vector for the line directly from its parameterization: $(1,0,-2)$. By inspection, two linearly independent vectors perpendicular to this are $(0,1,0)$ and $(2,0,1)$, and a point on the line can also be read directly from its parameterization, so, using the point-normal form of plane equation, the equations of two distinct planes that include the line are $y=1$ and $2x+z=2$. Every plane that includes the line has an equation that’s an affine combination of these two equations, yielding $$(1-\lambda)(y-1)+\lambda(2x+z-2) = 2\lambda x+(1-\lambda)y+\lambda z-(\lambda+1)=0.$$ Now use the distance formula with this generic plane equation and solve for $\lambda$.
A: Let $$ax+by+cz+d=0$$ the searched plane.Then it must be
$$a+b+d+ak-2kc=0$$ and
$$\left|\frac{-a+c+d}{\sqrt{a^2+b^2+c^2}}\right|=1$$
With our condition we get
$$|3c+d|=\sqrt{5c^2+b^2}$$
or by squaring
$$4c^2+d^2+6cd=b^2$$
with $$a+b+d=0$$ we can substitute $b$
$$b=-a-d$$
and we get
$$4c^2+d^2+6cd=a^2+d^2+2ad$$
or
$$cd=0$$
