Find the value of p, so that the lines l1: (1+x)/3 = 7y/p = z and l2: 7x/3p = y = (6-z)/5 are perpendicular to each other. Also find the equations of a line passing through a point (3, 2, 8) and parallel to line l1.

l1 = (1-x)/3 = 7y/p = z
l2 = 7x/3p = y = (6-z)/5
Given point P(3, 2, 8)
Direction cosines:
l1 = (3, p, 2)
l2 = (3p, 1, 5)
Given that both of the lines are ⊥. Therefore, (3, p, 2) . (3p, 1, 5) = 0
9p + p + 10 = 0
10p = -10
p = -1
Therefore direction cosine of line dc1 = (3, -1, 2)
Direction cosine of line l2 = (-7, 1, 5)
Required equation of line which is passing through P and parallel to l1 = (3, 2, 8) + λ(3, -1, 2).