CD = k1AB + k2CD
λi+3j-5k = k1(i-2j+2k) + k2(2i+j-3k)
=>
λ = k1+2k2 (1) and
3 = -2k1+ k2 (2) and
-5 = 2k1-3k2 (3)
(3) +(2)
-2k2=-2
k2 =1
k1 = -1
λ = k1+2k2 = -1+2 = 1