Let ${\mathrm{L}}_1:\frac{\mathrm{x}-5}{3}=\frac{\mathrm{y}-7}{-1}=\frac{\mathrm{z}+2}{1}$
${\mathrm{L}}_2:\frac{\mathrm{x}-3}{3}=\frac{\mathrm{y}-3}{2}=\frac{\mathrm{z}-6}{4}$
Be two lines in space. If  $L_3$ is a line with direction ratios <2, 7, - 5> meets L₁ and L₂ at A and B respectively, then the value of (AB)² is

$$\begin{gathered} \mathrm{A}(5+3\mathrm{\lambda },7-\mathrm{\lambda },-2+\mathrm{\lambda }) \\ \mathrm{B}(3+3\mathrm{\mu },3+2\mathrm{\mu },6+4\mathrm{\mu }) \end{gathered}$$
direction of $\mathrm{AB}=<2+3\mathrm{\lambda }-3\mathrm{\mu },4-\mathrm{\lambda }-2\mathrm{\mu },-8+\mathrm{\lambda }-4\mathrm{\mu }>$
AB is perpendicular to L₂
$11\mathrm{\lambda }-29\mathrm{\mu }-18=0 ...\left(1\right)$
Also  $\frac{2+3\mathrm{\lambda }-3\mathrm{\mu }}{2}=\frac{4-\mathrm{\lambda }-2\mathrm{\mu }}{7}=\frac{-8+\mathrm{\lambda }-4\mathrm{\mu }}{-5}$
$\Rightarrow \mathrm{\lambda }+\mathrm{m}+2=0 ...\left(2\right)$
from (1) and (2) $\mathrm{\lambda }=-1$
$\mathrm{\mu }=-1$
$\Rightarrow$A(2,8, -3) and B(0, 1, 2)
${\left(\mathrm{AB}\right)}^2=4+49+25=78$