Given the point $$P=(1, 2, 3)$$, find the distance between $$P$$ and the straight line $$r$$:
$$$r:(x,y,z)=(1,3,-2)+k\cdot(1,0,1)$$$
Development:
To find the distance to the straight line, we consider the point $$Q=(1, 3, -2)$$ and the governing vector $$\vec{v}= (1, 0, 1)$$. We calculate now the vector product of $$\overrightarrow{QP}$$ times $$\vec{v}$$.
$$\overrightarrow{QP}=(0,-1,5)$$
$$\begin{array}{rl} |\overrightarrow{QP}\times\vec{v}|=& \left| \begin{vmatrix} and & j & k \\ 0& -1& 5 \\ 1& 0& 1 \end{vmatrix} \right| = |-i+5j+k|=|(-3,2,-1)| \\ =&\sqrt{1+25+1}=\sqrt{27} \end{array}$$
And we can already apply the formula: $$$\text{d}(P,r)=\dfrac{|\overrightarrow{QP}\times\vec{v}|}{|\vec{v}|}= \dfrac{\sqrt{27}}{\sqrt{7}}=\sqrt{\dfrac{27}{2}}$$$
Solution:
$$\text{d}(P,r)=\sqrt{\dfrac{27}{2}}$$