Sphere with smallest radius 
Find the smallest sphere which touches the lines $\frac{x-5}{2}=\frac{y-2}{-1}=\frac{z-5}{-1}$ and $\frac{x+4}{-3}=\frac{y+5}{-6}=\frac{z-4}{4}$

The general equation of the sphere is $x^2+y^2+z^2+2ux+2vy+2wz+d=0$. Its centre is $(-u,-v,-w)$ and radius is $\sqrt(u^2+v^2+w^2-d)$. Since the given lines are tangent lines, the perpendicular distance from centre of sphere to them is equal to radius. If we equate them, we get two equations with 4 variables. I am wondering how to find the equation of sphere now?
Appreciate any hint.
 A: This video nicely explains a different approach based on geometric reasoning:
The two lines $L_1$ and $L_2$ are $<2t_1+5, 2-t_1,5-t_1>$ and $<-4-3t_2, -5-6t_2, 4+4t_2>$
The vector $\vec{P}$ between the two touchpoints $P_1$ and $P_2$ must be perpendicular to both direction vectors of the lines $L_1$ and $L_2$.
$$\begin{align}
\vec{P} = P_1 - P_2 & = \left[\array{(2t_1+5)-(-3t_2-4) \cr (2-t_)-(-6t_2-5) \cr (-t_1+5) - (4t_2+4)}\right] \\
& = \left[\array{2t_1+34t_2+9 \cr -t_1+6t_2+7 \cr -t_1-4t_2+1}\right] \\
\\
\vec{P}\left[\array{2 \cr -1 \cr -1}\right] & = 2(2t_1+3t_2+9) - 1(-t_1+6t_2+7) - 1(-t_1-4t_2+1) = 0 \\
& = 6t_1+4t_2+10=0\\
\\
\vec{P}\left[\array{-3 \cr -6 \cr 4}\right] & = -3(2t_1+3t_2+9) - 6(-t_1+6t_2+7) + 4(-t_1-4t_2+1) = 0 \\
& = 4t_1+61t_2+56=0 \\
\\
\\ 
t_1 = \frac{-784}{350}\\
t_2  = \frac{-296}{350} \\
&
\end{align}$$
$$\begin{align}
P_1 & = \{ 0.52 ; 4.24; 7.24 \} \\
P_2 & = \{  -1.463 ;  0.0743; .6171 \}
\end{align}$$
The center of the sphere is the middle between the touchpoints:
$$M = \{ -.4715; 2.1572; 3.929 \}$$
$$r^2 = \frac{|\vec{P}|}{4} = 15.1673$$
Thus the smallest sphere is the sphere with equation
$$(x+.4715)^2 + (y-2.1572)^2 + (z-3.929)^2 = 15.1673$$
