Find the values of p so that the lines $\cfrac{{1 - x}}{3} = \cfrac{{7y - 14}}{{2p}} = \cfrac{{z - 3}}{2}$
and $\cfrac{{7 - 7x}}{{3p}} = \cfrac{{y - 5}}{1} = \cfrac{{6 - z}}{5}$ are at right angles.

. : Equations of the given lines can be written as

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

Direction ratios of these lines are respectively

$< - 3,\cfrac{{2p}}{7},2 >$ and $< - \cfrac{{3p}}{7},1, - 5 >$

The lines are at right angles if
$( - 3) \times \left( {\cfrac{{ - 3p}}{7}} \right) + \left( {\cfrac{{2p}}{7}} \right) \times 1 + 2 \times ( - 5) = 0$

$\Rightarrow$ $\cfrac{{9p}}{7} + \cfrac{{2p}}{7} - 10 = 0 \Rightarrow \cfrac{{11p}}{7} = 10$ i.e. if $p = \cfrac{{70}}{{11}}$