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

DRS of both lines is equal to 0.

(3,2p,2)•(3p,1,5)=0

11p+10=0

p=-10/11