Distance between two lines in parametric form 
a) Parametrize the line $L$ through $P = (2, 1, 2)$ that intersects the
  line 
$x = 1 + t$, $y = 1 − t$, $z = 2t$ 
perpendicularly.
b) Parametrize the $z$ axis.
c) What is the distance from this line $L$ to the $z$-axis?

My work For the part a), I got the equation was $x = 2 + \frac{t}{6}$ ; $y = 1 + \frac{5t}{6}$ ; $z = 2 + \frac{t}{3}$. 
Stuck with the last 2 parts. Anyone want to give this math novice a hand? :D 
 A: A parametrization of the $z$ axis is $x=0$, $y=0$, $z=s$ with $s\in\mathbb{R}$. 
Therefore the squared distance between $P$ along the line $L$,  and $Q$ along the $z$-axis, is the $2$-variable function 
$$f(t,s)=(2 + (1/6)t-0)^2+(1 + (5/6)t-0)^2+( 2 + (1/3)t-s)^2.$$
Note that the minimum distance will be attained when the partial derivatives are zero: 
$$f_t(t,s)=\frac{-2t+6s-12}{3}=0\quad f_s=\frac{5t-2s+11}{3}=0\implies t=-\frac{21}{13}\; s =\frac{19}{13}$$ 
Therefore the distance between the line $L$ and the $z$ axis is
$$\sqrt{f(-21/13,19/13)}=\frac{9}{\sqrt{26}}.$$
A: Let $B(1+t,1-t,2t)$ and $A(2,1,2)$.
Thus, $$\vec{AB}\perp\vec{(1,-1,2)}$$ or
$$\vec{(t-1,-t,2t-2)}\vec{(1,-1,2)}=0$$ or
$$t-1+t+4t-4=0$$ or
$$t=\frac{5}{6}$$ and since
$$\vec{\left(\frac{5}{6}-1,-\frac{5}{6},\frac{5}{3}-2\right)}=\vec{\left(-\frac{1}{6},-\frac{5}{6},-\frac{1}{3}\right)},$$
we obtain:
$$L:(2,1,2)+s(1,5,2),$$
which you got by yourself.
Now, let $\pi$ be a plain in which placed $z$-axis such that $\pi||L.$
Thus, the parametric equation of $\pi$ it's
$$(x,y,z)=(0,0,0)+t(0,0,1)+s(1,5,2).$$
Let $\vec{n}(a,b,c)$ be a normal of $\pi$.
Thus,
$$0\cdot a+0\cdot b+1\cdot c=0$$ and
$$a+5b+2c=0,$$
which gives that we can assume that $\vec{n}(5,-1,0)$ and we got an equation of the plain $\pi$:
$$5(x-0)-1(y-0)+0(z-0)=0$$ or
$$5x-y=0.$$
Now, we can calculate our distance:
$$\frac{|5\cdot2-1\cdot1+0\cdot2|}{\sqrt{5^2+1^2+0^2}}=\frac{9}{\sqrt{26}}.$$
