Find the shortest distance between the lines

$\cfrac{{x + 1}}{7} = \cfrac{{y + 1}}{{ - 6}} = \cfrac{{z + 1}}{1}$ and $\cfrac{{x - 3}}{1} = \cfrac{{y - 5}}{{ - 2}} = \cfrac{{z - 7}}{1}$

.: The shortest distance between the lines
${l_1}:\cfrac{{x - {x_1}}}{{{a_1}}} = \cfrac{{y - {y_1}}}{{{b_1}}} = \cfrac{{z - {z_1}}}{{{c_1}}}$

and ${l_2}:\cfrac{{x - {x_2}}}{{{a_2}}} = \cfrac{{y - {y_2}}}{{{b_2}}} = \cfrac{{z - {z_2}}}{{{c_2}}}$

is $\left| {\cfrac{{\left| {\begin{array}{cccccccccccccccccccc}{{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}}\\{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\end{array}} \right|}}{{\sqrt {{{({b_1}{c_2} - {b_2}{c_1})}^2} + {{({c_1}{a_2} - {a_1}{c_2})}^2} + {{({a_1}{b_2} - {a_2}{b_1})}^2}} }}} \right|$

Here, ${x_1} = - 1,{y_1} = - 1,{z_1} = - 1;{x_2} = 3,{y_2} = 5,{z_2} = 7$

and ${a_1} = 7,{b_1} = - 6,{c_1} = 1;{a_2} = 1,{b_2} = - 2,{c_2} = 1$

Required shortest distance
$= \left| {\cfrac{{\left| {\begin{array}{cccccccccccccccccccc}4&6&8\\7&{ - 6}&1\\1&{ - 2}&1\end{array}} \right|}}{{\sqrt {{{[ - 6 - ( - 2)]}^2} + {{(1 - 7)}^2} + {{[ - 14 - ( - 6)]}^2}} }}} \right|$

$= \left| {\cfrac{{4( - 6 + 2) - 6(7 - 1) + 8( - 14 + 6)}}{{\sqrt {16 + 36 + 64} }}} \right| = \left| {\cfrac{{4( - 4) - 6(6) + 8( - 8)}}{{\sqrt {116} }}} \right|$

$= \cfrac{{116}}{{\sqrt {116} }} = \sqrt {116} = 2\sqrt {29}$ units.