Maybe the following will help. It's not an answer, but it's too long for a comment.
Suppose the given distance is $d$. Then $L$ must be simultaneously tangent to three spheres of radius $d$, one centered at each of the given points.
Consider the case where the distances between some pair of the points is $D>2d$. The radius $d$ spheres can be nestled inside a ruled surface of this shape (so long as it's sized appropriately):
Any line of the ruled surface is tangent to both spheres.
You can also fit two spheres with centers distance $D$ apart into a ruled surface that's more (or completely) cone-like, or into one that's less cone-like and more (or exactly) cylindrical. Any of these ruled surfaces will yield an infinite number of lines $L$ tangent to these first two spheres.
If the sphere of radius $d$ about the third point intersects any of these ruled surfaces (hyperboloids of one sheet with axis containing the first two spheres' centers - the cone and the cylinder being degenerate ones), it seems like there will be either one or two lines $L$ from that ruled surface where $L$ equidistant from the three points.
I think that the third sphere will intersect one of these ruled surfaces if and only if it intersects either the cone or the cylinder, so maybe only the degenerate hyperboloids need to be considered.
If none of the points are more than $2d$ units apart, the first two spheres will intersect in a circle. The families of lines tangent to both will comprise a cylinder and some hyperboloids, but you can't twist them completely into a cone. Instead, there will be a different degenerate ruled family of tangent lines: those in the plane through the spheres' circle of intersection that are also tangent to that circle. Again, if the third sphere intersects any of these families of lines (and I think in this case it must), you'll get solutions.
Vectors is a fine way to solve this. Find make $(1,2,-1)$ the tail. Make (1,-1,1) the direction, and find $t$ such that $(1,2,-1) + t\frac {(1,-1,1)}{\|(1,-1,1)\|} = 0$
There is also a distance formula:
What does the old fashioned distance formula say.
$d = \frac {|ax + by + cz - d|}{\sqrt {a^2 + b^2 + c^2}} = \frac {5}{\sqrt {3}}$
Lagrange multiplier are entirely unnecessary in this problem. But they are useful in more complicated optimization problems.
If you want to use Lagrange multipliers:
$d = \sqrt {(x-1)^2 + (y-2)^2 + (z+1)^2}$ is minimized when $d^2 = (x-1)^2 + (y-2)^2 + (z+1)^2$
We can say this because when $a,b > 0$, and $a>b$ then $a^2 > b^2$
minimizing $d^2$ will minimize $d$
$\frac {\partial F}{\partial x} = 2(x-1) - \lambda = 0\\
\frac {\partial F}{\partial y} = 2(y-2) + \lambda = 0\\
\frac {\partial F}{\partial z} = 2(z+1) - \lambda = 0$
$x = \frac {\lambda + 2}{2}\\
y = \frac {-\lambda + 4}{2}\\
z = \frac {\lambda - 2}{2}$
plug into the constraint (mutiplying by 2 to kill the deominator):
$2x - 2y + 2z = 3\\
\lambda + 2 + \lambda - 4 + \lambda - 2 = 6\\
\lambda = \frac {10}{3}$
$(x,y,z) = (\frac {8}{3}, \frac {1}{3}, \frac {13}{6})$
$d = \sqrt {\frac {5}{3}^2 + \frac {-5}{3}^2 + \frac {5}{3}^2} = \frac {5\sqrt 3}{3}$
Best Answer
Consider the Lagrange function $$L(\mathbf x,\lambda)=\|\mathbf x-\mathbf x_0\|^2+2\lambda(\mathbf w^T\mathbf x+b)$$ The Lagrange multiplier is multiplied by $\,2\,$ to simplify the computations (this is legal).
Since $$L(\mathbf x,\lambda)=\|\mathbf x\|^2-2(\mathbf x_0-\lambda \mathbf w)^T\mathbf x+2\lambda b+\|\mathbf x_0\|^2$$ one has $$\frac {\partial L}{\partial \mathbf x}=2\mathbf x-2(\mathbf x_0-\lambda \mathbf w)$$ using formulas (69) and (131) in the pdf quoted by @user25004.
Solving $\,\dfrac {\partial L}{\partial \mathbf x}=0\,$, one obtains $$\mathbf x=\mathbf x_0-\lambda \mathbf w$$ which, substituted in the equation of the hyperplane, gives $$\lambda=\frac {\mathbf w^T\mathbf x_0+b}{\|\mathbf w\|^2}$$ so the shortest distance is $$\|\mathbf x_0-\lambda \mathbf w-\mathbf x_0\|=|\lambda|\|\mathbf w\|=\frac {|\mathbf w^T\mathbf x_0+b|}{\|\mathbf w\|}$$