Equation of plane containing the line and lying at a distance from the point

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 $\lambda$.