Consider two lines ${\mathrm{L}}_1 : \frac{\mathrm{x}-7}{3}=\frac{\mathrm{y}-7}{2}=\frac{\mathrm{z}-3}{1}\text{ and }{\mathrm{L}}_2 : \frac{\mathrm{x}-1}{3}=\frac{\mathrm{y}+1}{4}=\frac{\mathrm{z}+1}{3}$. If a line L whose direction ratios are < 2,2,1 > interest the line ${\mathrm{L}}_1\text{ and }{\mathrm{L}}_2$ at A and B then the distance of AB is_____

$$\begin{gathered} Point on the line \\ {\mathrm{L}}_1 : \frac{\mathrm{x}-7}{3}=\frac{\mathrm{y}-7}{2}=\frac{\mathrm{z}-3}{1}=\lambda \end{gathered}$$

$\begin{array}{l}\mathrm{A}(3\mathrm{\lambda }+7, 2\mathrm{\lambda }+7, \mathrm{\lambda }+3); \\ Point on the line{\mathrm{L}}_2 : \frac{\mathrm{x}-1}{3}=\frac{\mathrm{y}+1}{4}=\frac{\mathrm{z}+1}{3}=\mu \\ \mathrm{B}(2\mathrm{\mu }+1, 4\mathrm{\mu }-1, 3\mathrm{\mu }-1)\\ \frac{3\mathrm{\lambda }-2\mathrm{\mu }+6}{2}=\frac{2\mathrm{\lambda }-4\mathrm{\mu }+8}{2}=\frac{\mathrm{\lambda }-3\mathrm{\mu }+4}{1}\end{array}$

On solving we get $\mathrm{\lambda }=2\text{ and }\mathrm{\mu }=0$

Hence A(13,11,5) B(1,–1,–1)

$\mathrm{AB}=\sqrt{144+144+36}=18$