The distance between two straight lines $$r$$ and $$r'$$, $$\text{d}(r,r')$$, is the minimal distance between any point of $$r$$ and any other point of $$r'$$.
 If the straight lines coincide or are secant, the distance between them is zero, $$\text{d}(r,r')=0$$.
 If the straight lines are parallel, the distance between them can be calculated at any point of one of the lines, $$P\in r$$ or $$P'\in r'$$, and finding the distance to the other straight line: $$\text{d}(r,r')=\text{d}(P,r')=\text{d}(r,P')$$

If the straight lines cross, the following general formula is deduced to calculate the distance between them:
We take a point $$A$$ belonging to $$r$$ and another point $$A'$$ belonging to $$r'$$. Let $$\vec{v}$$ and $$\vec{v}'$$ be the governing vectors of $$r$$ and $$r'$$. We join the points $$A$$ and $$A'$$. The volume of the parallelepiped determined by $$\overrightarrow{AA'}$$, $$\vec{v}$$ and $$\vec{v}'$$, is the absolute value of the mixed product of these vectors: $$$v_p=[\overrightarrow{AA'},\vec{v},\vec{v}']$$$
On the other hand we can also calculate this volume by multiplying the area of the base and the height: $$$v_p=\vec{v}\times\vec{v}'\text{d}(r,r')$$$
Therefore: $$$\text{d}(r,r')=\dfrac{[\overrightarrow{AA'},\vec{v},\vec{v}']} {\vec{v}\times\vec{v}'}$$$
We are going to calculate the distance between the straight lines: $$$ r:x2=\dfrac{y+3}{2}=z \qquad r':x=y=z$$$
First we determine its relative position. To do it we must write the implicit equations of the straight line: $$$ r:\left\{ \begin{array}{l} 2xy7=0 \\ xz2=0 \end{array} \right. \qquad r':\left\{ \begin{array}{l} xy=0 \\ xz=0 \end{array} \right.$$$
And we calculate the rank of the matrix of the resulting systems of equations: $$$M'=\begin{vmatrix} 2 & 1 & 0 & 7 \\ 1 & 0 & 1 & 2 \\ 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \end{vmatrix} =2 \neq 0 $$$
Therefore $$\text{rank}(M')=4$$ and the two straight lines intersect. Now we must find a point and the governing vector of each line.
For the straight line $$r$$: $$A=(2,3,0)$$ and $$\vec{v}=(1,2,1)$$.
For the straight line $$r'$$: $$A'=(0,0,0)$$ and $$\vec{v}=(1,1,1)$$.
So we have: $$\overrightarrow{AA'}=(2,3,0)$$
$$$\begin{array}{rl} \vec{v}\times\vec{v}'=&\left \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 2 & 1 \\ 1 & 1 & 1 \end{vmatrix} \right= 2\vec{i}+\vec{j}+\vec{k}2\vec{k}\vec{j}\vec{i}= \vec{i}\vec{k} \\ =& (1,0,1) = \sqrt{1^2+0^2+(1)^2}=\sqrt{2} \end{array}$$$
$$$[\overrightarrow{AA'},\vec{v},\vec{v}']= \begin{vmatrix} 2 & 3 & 0 \\ 1 & 2 & 1 \\ 1 & 1 & 1 \end{vmatrix} = 4+3+23=2 $$$
Finally: $$$ \text{d}(r,r')=\dfrac{[\overrightarrow{AA'},\vec{v},\vec{v}']} {\vec{v}\times\vec{v}'}= \dfrac{2}{\sqrt{2}}=\sqrt{2}$$$