a For L1 v1 = 3i - 2j + 6k

L2 v2 = i - 3j + 2k

the angle q between the lines is the angle between v1 and v2:

cos(q) = v1*v2/(|v1||v2|)

b

set the x coordintates equal 3t - 3 = s -6

s = 3t+3

set the y components equal -2t = -3s -5 = -9t - 9 +5

At the pt of intersection t = -2 and s =-3 yielding (-9,4,-5) as the pt of int

c. The normal to the plane is v1 x v2 = N = ai + bj + ck. choose a pt on either line , say (-3,0,7)

Use a(x+3) +by + c(z-7) = 0 to obtain eqn of plane