Find the values `p` so that line `(1-x)/3=(7y-14)/(2p)=(z-3)/2a n d(7-7x)/(3p)=(y-5)/1=(6-z)/5` are at right angles.

Explanation

The given equations can be written in the standard form as
`" "(x-1)/(-3)=(y-2)/(2p//7)=(z-3)/(2) and (x-1)/(-3p//7)=(y-5)/(1)=(z-6)/(-5)`
The direction ratios of the lines are `-3, (2p)/(7),2 and (-3p)/(7), 1, -5, ` respectively.
Lines are perpendicular to each other. Therefore,
`" "(-3)*((-3p)/(7))+((2p)/(7))*(1)+2*(-5)=0`
or `" "(9p)/(7)+(2p)/(7)=10 or 11p=70 or p=(70)/(11)`